JEE-PhysicsNode-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 E 113Various parts of electromagnetic spectrum S. Radiation Discover How Wavelength Range Frequency range Energy Properties Application No. produced range 3×1022Hz to 107eV-104 eV 1. -Rays Henry Due to decay 10–14m to 10–10m 3×1018 (a) High (a)Gives Becquerel of radioactive 2.4×105eV to and nuclei. 1.2×103eV penetrating Information on Madam Cuire power nuclear structure (b) Uncharged (b) Medical trea- (c) Low ionising tment etc. power 2. X-Ray Roentgen Due to collisions 6×10–12m to 10–9m 5×1019 Hz to (a)Low Penetrating (a) Medical of high energy 3×1017 Hz electrons with power diagnosis and heavy targets (b) other properties treatment similar to -rays (b) Study of except wavelength crystal structure (c) Industrial radiography 3. Ultraviolet Ritter By ionised gases, 6×10–10m to 3×1017 Hz to 2×103eV to 3eV (a) All properties (a) To detect Rays sun lamp 3.8×10–7m 5×1019 Hz spark etc. of light adulteration, (b) Photoelectric writing and effect signature (b) Sterilization of water due to its destructive action on bacteria 4. Visible light Newton Outer orbit electron 3.8×10–7 m to 8×1014Hz to 3.2 eV to 1.6 eV (a) Sensitive to (a)To see objects transitions in atoms, 7.8×10–7m 4×1014 Hz Subparts of gas discharge tube, human eye (b) To study visible incandescent solids spectrum and liquids. molecular (a) Violet (b) Blue structure (c) Green (d) Yellow 3.9×10–7m to 4.55×10–7m 7.69×1014Hz to 6.59×1014Hz (e) Orange 4.55×10–7m to 4.92×10–7m 6.59×1014Hz to 6.10×1014Hz (f) Red 4.92×10–7m to 5.77×10–7m 6.10×1014Hz to 5.20×1014Hz 5.77×10–7m to 5.97×10–7m 5.20×1014Hz to 5.03×1014Hz 5.97×10–7m to 6.22×10–7m 5.03×1014Hz to 4.82×1014Hz 6.22×10–7m to 7.80×10–7m 4.82×1014Hz to 3.84×1014Hz
JEE-PhysicsS. Radiation Discover HowWavelength Range Frequency rangeEnergyPropertiesApplication No. producted range 114 E 5. Infra-Red Willam (a) Rearrangement 7.8×10–7m to 10–3m 4×1014Hz to 3×1011Hz 1.6eV to 10–3eV (a) Thermal effect (a) Used in indu- 3×1011 Hz to 109 Hz (b) All properties stry, medicine waves Herschell of outer orbital simillar to those of and astronomy light except (b) Used for fog electrons in atoms or haze photography (c) Elucidating and molecules. molecular structure. (b) ChangeE of molecular vibrational and rotational energies (c) By bodies at high temperature. 6. Microwaves Hertz Special electronic 10–3 to 0.3m 10–3ev to 10–5eV (a) Phenomena of (a) Radar and teleco- 10–3eV to 0 reflection, mmunication. devices such as refraction and (b) Analysis of fine diffraction details of molecular klystron tube structure (a) Exhibit waves (a) Radio 7. Radio waves Marconi Oscillating circuits 0.3 to few kms. 109Hz to few Hz like properties communication more than particle like properties. Subparts of 0.01m to 0.1m 3×1010Hz to 3×109Hz Radar, Radio and satelite communication Radio- 0.1 m to 1m 3×109Hz to 3×108Hz (Microwaves), Radar and Television spectrum 1 m to 10 m 3×108Hz to 3×107Hz broadcast short distance communication, (A) Super High Frequency 10m to 100 m 3×107Hz to 3×106Hz Television communication. (a) SHF 100 m to 1000 m 3×106Hz to 3×105Hz Ultra High Frequency 1000 m to 10000 m 3×105Hz to 3×104Hz Medium distance communication (b) UHF 10000 m to 30000 m 3×104Hz to 104Hz Telephone communication, Marine and Very High Frequency navigation use, long range communi- (c) VHF cation. Long distance communication. (B) High Frequency (HF) Medium Frequency (MF) Low Frequency (LF) Very Low Frequency (VLF) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65
JEE-Physics SOME WORKED OUT EXAMPLES Example#1 In a plane electomagnetic wave, the electric field oscillates sinusoidally at a frequency of 2 × 1010 Hz and amplitude 48 V/m. The amplitude of oscillating magnetic field will be – Solution Oscillating magnetic field B E 48 16 108 Wb/m2 c 3 108 Example#2 In the above problem, the wavelength of the wave will be – Solution Wavelength of electromagnetic wave c 3 108 1.5 102 1.5 cm 2 1010 Example#3 A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5 m from the source will be – Solution I Pav E 2 4 r 2 m Intensity of electromagnetic wave given is by 2oc o cPav (4 107 ) (3 108 ) 800 2 r 2 2 3.5 2 Em = = 62.6 V/m Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 Example#4 In the above problem, the maximum value of magnetic field will be – Solution The maximum value of the magnetic field is given by Bm Em 62.6 2.09 × 10–7 T c = 3 108 Example#5 What should be the height of transmitting antenna if the T.V. telecast is to cover a radius of 128 km ? Solution Height of transmitting antenna h d2 (128 103 )2 1280 m 2Re 2 6 4 106 E 115
JEE-Physics Example#6 The area to be covered for T.V. telecast is doubled, then the height of transmitting antenna (T.V. tower) will have to be – Solution The area of transmission of surrounding the T.V. tower A = d2 = (2hRe) A h Example#7 In an electomagnetic wave, the amplitude of electric field is I V/m. The frequency of wave is 5 × 1014 Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be Solution Average energy dendity is given by uE 1 o E 2 1 o Eo 2 1 o E 2 1 8.85 1012 (1)2 = 2.2 × 10–12 J/m2 2 2 2 4 o 4 Example#8 A T.V. tower has a height of 100 m. How much population is covered by T.V. broadcast, if the average population density around the tower is 1000/km2 ? Solution Radius of the area coved by T.V. telecast d 2hR e Total population covered = d2 × population density = 2hRe × polulation density 1000 = 2 × 3.14 × 100 × 6.4 × 106 × 106 = 39.503 × 105 Example#9 An electomagnetic radiation has an energy 14.4 KeV. To which region of electromagnetic spectum does it belong ? Solution hc 6.6 1034 3 108 = 0.8 × 10–10 m = 0.8Å 14.4 103 1.6 1019 E This wavelength belongs to X–ray region Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 116 E
JEE-Physics EXERCISE 1 CHECK YOUR GRASP 1 . If E and B are the electric and magnetic field vectors of electromagnetic waves then the direction of propagation of electromagnetic wave is along the direction of – (4) none of these (1) E (2) B (3) E × B 2 . The electromagnetic waves do not transport - (1) energy (2) charge (3) momentum (4) information 3 . In an electromagnetic wave the average energy density is associated with – (1) electric field only (2) magnetic field only (3) equally with electric and magnetic fields (4) average energy density is zero 4 . In an electromagnetic wave the average energy density associated with magnetic field will be (1) 1 L I2 B2 (3) 1 0 B 2 (4) 1 0 2 (2) 20 2 2 B2 5 . In the above problem, the energy density associated with the electric field will be – (1) 1 CV 2 1 q2 1 2 (4) 1 0 E 2 2 (2) (3) 2 2C 2E 6 . In which part of earth's atmosphere is the ozone layer present ? (1) troposphere (2) stratosphere (3) ionosphere (4) meosphere 7 . The ozone layer is earth's atmosphere is crucial for human survival because it – (1) has ions (2) reflects radio signals (3) reflects ultraviolet rays (4) reflects infra red rays Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 8 . The frequency from 3 × 109 Hz to 3 × 1010 Hz is – (2) super high frequency band (1) high frequency band (4) very high frequency band (3) ultra high frequency band 9 . The frequency from 3 to 30 MHz is known as – (2) medium frequency band (1) audio band (4) high frequency band (3) very high frequency band 1 0 . The AM range of radiowaves have frequency – (1) less than 30 MHz (2) more than 30 MHz (3) less than 20000Hz (4) more than 20000Hz 11. Select wrong statement from the following for EMW - (2) travel with same speed in all medium (1) are transverse (4) are produced by accelerating charge E (3) travel with the speed of light 117
JEE-Physics (3) transverse (4) transverse stationary 1 2 . The nature of electromagnetic wave is – (1) longitudinal (2) longitudinal stationary 1 3 . Which of the following are not electromagnetic waves ? [AIEEE–2002] (1) Cosmic-rays (2) -rays (3) -rays (4) X-rays 1 4 . The rms value of the electric field of the light coming from the sun is 720 N/C. The average total energy density of the electromagnetic wave is- [AIEEE–2006] (1) 4.58 × 10–6 J/m3 (2) 6.37 × 10–9 J/m3 (3) 81.35 × 10–12 J/m3 (4) 3.3 × 10–3 J/m3 15. An electromagnetic wave in vacuum has the electric and magnetic fields E and B, which are always perpendicular to each other. The direction of polarizaiton is given by X and that of wave propagation by k . Then :- [AIEEE - 2012] (2) and k||B E (1) X | | E and k| | B E X | | B (3) X| | E and k| | E B (4) X| | B and k| | E B 1 6 . An electromagnetic wave with frequency and wavelength travels in the +y direction. Its magnetic field is along _ x axis. The vector equation for the associated electric field (of amplitude E0) is :- [AIEEE–2012 (Online)] (1) E0 cos t 2 y xˆ (2) E 0 cos t 2 y xˆ E E (3) E 0 cos t 2 y ˆz (4) E0 cos t 2 y ˆz E E ANSWER KEY EXERCISE Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 A 3 2 3 2 4 2 3 2 2 1 2 3 1, 3 1 3 4 118 E
JEE-Physics ELECTROSTATICS ELECTRIC CHARGE Charge is the property associated with matter due to which it produces and experiences electrical and magnetic effects. The excess or deficiency of electrons in a body gives the concept of charge. Types of charge : (i) Positive charge : It is the deficiency of electrons as compared to proton. (i) Negative charge : It is the excess of electrons as compared to proton. SI unit of charge : ampere × second i.e. Coulomb Dimension : [A T] Practical units of charge are ampere × hour (=3600 C) and faraday (= 96500 C) • Millikan calculated quanta of charge by 'Highest common factor' (H.C.F.) method and it is equal to charge of electron. • 1 C = 3 109 stat coulomb, 1 absolute - coulomb = 10 C, 1 Faraday = 96500 C. SPECIFIC PROPERTIES OF CHARGE • Charge is a scalar quantity : It represents excess or deficiency of electrons. • Charge is transferable : If a charged body is put in contact with an another body, then charge can be transferred to another body. • Charge is always associated with mass Charge cannot exist without mass though mass can exist without charge. • So the presence of charge itself is a convincing proof of existence of mass. • In charging, the mass of a body changes. • When body is given positive charge, its mass decreases. • When body is given negative charge, its mass increases. • Charge is quantised The quantization of electric charge is the property by virtue of which all free charges are integral multiple of a basic unit of charge represented by e. Thus charge q of a body is always given by q = ne n = positive integer or negative integer The quantum of charge is the charge that an electron or proton carries. Note : Charge on a proton = (–) charge on an electron = 1.6 × 10–19 C • Charge is conserved In an isolated system, total charge does not change with time, though individual charge may change i.e. charge can neither be created nor destroyed. Conservation of charge is also found to hold good in all types of reactions either chemical (atomic) or nuclear. No exceptions to the rule have ever been found. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 • Charge is invariant Charge is independent of frame of reference. i.e. charge on a body does not change whatever be its speed. Accelerated charge radiates energy v = 0 (i.e. at rest) v = constant v constant (i.e. time varying) + + + Q Q Q produces only E produces both E and B produces E, B (electric field) (magnetic field) but no radiation and radiates energy • Attraction – Repulsion Similar charges repel each other while dissimilar attract E1
JEE-Physics METHODS OF CHARGING • F ri ct i on : If we rub one body with other body, electrons are transferred from one body to the other. Transfer of electrons takes places from lower work function body to higher work function body. Positive charge Negative charge Glass rod Silk cloth Woollen cloth Rubber shoes, Amber, Plastic objects Dry hair Comb Flannel or cat skin Ebonite rod Note : Clouds become charged by friction • Electrostatic induction If a charged body is brought near a metallic neutral body, the charged body will attract opposite charge and repel similar charge present in the neutral body. As a result of this one side of the neutral body becomes negative while the other positive, this process is called 'electrostatic induction'. • Charging a body by induction (in four successive steps) charging q'=0 charging q'=-ve charging q'=-ve q'=-ve body body body charged body is uncharged body is uncharged body is charging brought near connected to disconnected body earth from the earth uncharged body step-2 step-3 is removed step-4 step-1 Some important facts associated with induction- (i) Inducing body neither gains nor loses charge (ii) The nature of induced charge is always opposite to that of inducing charge (iii) Induction takes place only in bodies (either conducting or non conducting) and not in particles. • Conduction The process of transfer of charge by contact of two bodies is known as conduction. If a charged body is put in NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 contact with uncharged body, the uncharged body becomes charged due to transfer of electrons from one body to the other. • The charged body loses some of its charge (which is equal to the charge gained by the uncharged body) • The charge gained by the uncharged body is always lesser than initial charge present on the charged body. • Flow of charge depends upon the potential difference of both bodies. [No potential difference No conduction]. • Positive charge flows from higher potential to lower potential, while negative charge flows from lower to higher potential. 2 E
JEE-Physics GOLDEN KEY POINTS • Charge differs from mass in the following sense. (i) In SI units, charge is a derived physical quantity while mass is fundamental quantity. (ii) Charge is always conserved but mass is not (Note : Mass can be converted into energy E=mc2 (iii) The quanta of charge is electronic charge while that of mass it is yet not clear. (iv) For a moving charged body mass increases while charge remains constant. • True test of electrification is repulsion and not attraction as attraction may also take place between a charged and an uncharged body and also between two similarly charged bodies. q • For a non relativistic (i.e. v << c) charged particle, specific charge =constant m q • For a relativistic charged particle decreases as v increases, where v is speed of charged body. m Example When a piece of polythene is rubbed with wool, a charge of – 2 × 10–7 C is developed on polythene. What is the amount of mass, which is transferred to polythene. Solution Q 2 107 From Q = ne, So, the number of electrons transferred n e = 1.6 1019 = 1.25 × 1012 Now mass of transferred electrons = n × mass of one electron = 1.25 × 1012 × 9.1 × 10–31 = 11.38 × 10–19 kg Example 1012 – particles (Nuclei of helium) per second falls on a neutral sphere, calculate time in which sphere gets charged by 2C. Solution Number of – particles falling in t second = 1012t Charge on – particle = +2e , So charge incident in time t = (1012t).(2e) Given charge is 2 C 10 18 2 × 10–6 = (1012t).(2e) t 1.6 1019 6.25s NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 COULOMB'S LAW The electrostatic force of interaction between two static point electric charges is directly proportional to the product of the charges, inversely proportional to the square of the distance between them and acts along the straight line joining the two charges. If two points charges q and q separated by a distance r. Let F be the electrostatic force between these two 12 charges. According to Coulomb's law. F q q and 1 + –+ + 1 2 r2 F q2 on1 1 q2 F q1on 2 2 F F1 on q1 F2on1 2 kq1q2 1 9 109 Nm2 r2 k 4 0 k = coulomb's constant or electrostatic force constant F = where C2 e E3
JEE-Physics Coulomb's law in vector form = force on q due to q = kq1q2 ˆr21 F12 r F21 12 r2 q1 r12 q2 F12 = kq1q2 ˆr12 (here ˆr12 is unit vector from q to q ) r2 1 2 F21 Coulomb's law in terms of position vector y r21 q2 q1 r2 x = kq 1 q 2 ( r1 r2 ) r1 F12 r1 r2 3 O Principle of superposition The force is a two body interaction, i.e., electrical force between two point charges is independent of presence or absence of other charges and so the principle of superposition is valid, i.e., force on a charged particle due to number of point charges is the resultant of forces due to individual point charges, i.e., F1 = F12 + F13 +... Note : Nuclear force is many body interaction, so principle of superposition is not valid in case of nuclear force. When a number of charges are interacting, the total force on a given charge is vector sum of the forces exerted on it by all other charges individually kq0q1 kq0q2 ... kq0qi ... kq0qn n qi r12 r22 ri2 rn2 i 1 ri2 in vector form F kq0 F ˆri SOME IMPORTANT POINTS REGARDING COULOMB’S LAW AND ELECTRIC FORCE • The law is based on physical observations and is not logically derivable from any other concept. Experiments till today reveal its universal nature. • The law is analogous to Newton’s law of gravitation : F = G m1m2 with the difference that : r2 (a) Electric force between charged particles is much stronger than gravitational force, i.e., F >>F . This is EG why when both F and F are present, we neglect F . EG G (b) Electric force can be attractive or repulsive while gravitational force is always attractive. (c) Electric force depends on the nature of medium between the charges while gravitational force does not. (d) The force is an action–reaction pair, i.e., the force which one charge exerts on the other is equal and opposite to the force which the other charge exerts on the first. • The force is conservative, i.e., work done in moving a point charge once round a closed path under the action of Coulomb’s force is zero. • The net Coulomb’s force on two charged particles in free space and in a medium filled upto infinity are NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 1 q1q2 and F' = 1 q1q2 . So F = K, F = 40 r2 4 r2 = 0 F' Dielectric constant (K) of a medium is numerically equal to the ratio of the force on two point charges in free space to that in the medium filled upto infinity. • The law expresses the force between two point charges at rest. In applying it to the case of extended bodies of finite size care should be taken in assuming the whole charge of a body to be concentrated at its ‘centre’ as this is true only for spherical charged body, that too for external point. Although net electric force on both particles change in the presence of dielectric but force due to one charge particle on another charge particle does not depend on the medium between them. • Electric force between two charges does not depend on neighbouring charges. 4E
JEE-Physics Example If the distance between two equal point charges is doubled and their individual charges are also doubled, what would happen to the force between them? Solution 1 qq 1 (2q) (2q) 1 4q2 1 q2 F = 40 r2 ....(i) 4r2 = 40 r2 Again, F' = 40 (2r )2 or F' = = F 4 0 So, the force will remain the same. Example A particle of mass m carrying charge '+q ' is revolving around a fixed charge '–q ' in a circular path of 12 radius r. Calculate the period of revolution. Solution 1 q1q2 = mr2 = 42mr 40 r2 T2 T2 = (40 )r 2(42mr) or 0mr q1q2 T = 4r q1q2 where is the vector drawn from source charge is test charge. r Example The force of repulsion between two point charges is F, when these are at a distance of 1 m. Now the point charges are replaced by spheres of radii 25 cm having the charge same as that of point charges. The distance between their centres is 1 m, then compare the force of repulsion in two cases. Solution In 2nd case due to mutual repulsion, the effective distance between their centre of charges will be increased (d' > d) so force of repulsion decreases as F 1 d2 d d' EQUILIBRIUM OF CHARGED PARTICLES NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 In equilibrium net electric force on every charged particle is zero. The equilibrium of a charged particle, under the action of Colombian forces alone can never be stable. Equilibrium of three point charges (i) Two charges must be of like nature as Fq KQ1q KQ2q 0 x Q2 x2 Q1 q r x2 r (ii) Third charge should be of unlike nature as FQ1 KQ1Q2 KQ1q 0 r2 x2 Therefore x Q1 r and q = Q1 Q 2 ( Q1 Q2 )2 Q1 Q2 E5
JEE-Physics Equilibrium of symmetric geometrical point charged system Value of Q at centre for which system to be in state of equilibrium q q a q aa a Qa aq Q q a qq q q(2 2 1) (i) For equilateral triangle Q = 3 (ii) For square Q = 4 Equilibrium of suspended point charge system For equilibrium position Tcos = mg and sin F = kQ2 Fe kQ2 T = e tan = mg = x2mg Tcos T x2 Fe Q Q • If is small then tan Tsin x mg x x kQ2 1 ~~ sin = 2 2 = x2 mg x3 = 2kQ2 Q2 3 x mg 2 0 mg • If whole set up is taken into an artificial satellite (g ~ 0) then T = F= kq2 2 eff e 180° 4 2 q q Example q a Q For the system shown in figure find Q for which resultant force on q is zero. (0,a) (a,a) Solution a FA For force on q to be zero, charges q and Q must be of opposite of nature. q Net attraction force on q due to charges Q = Repulsion force on q due to q (0,0) FR Q a FA kQq kq2 F = F 2 = ( 2a )2 q = 2 2 Q Hence q = – 2 2Q 2 AR a2 Example NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Two identically charged spheres are suspended by strings of equal length. The strings make an angle of 300 with each other. When suspended in a liquid of density 0.8 g/cc the angle remains same. What is the dielectric constant of liquid. Density of sphere = 1.6 g/cc. 300 E 150 F mg 6
JEE-Physics Solution F When set up shown in figure is in air, we have tan 150 = mg When set up is immersed in the medium as shown F in figure, the electric force experienced by the ball will reduce and will be equal to r and the effective gravitational force will become mg Thus we have tan 150 = F F1 =2 1 s mg r mg r 1 1 s s 300 15° F r mgeff=mg 1 Example Given a cube with point charges q on each of its vertices. Calculate the force exerted on any of the charges due to rest of the 7 charges. Solution The net force on particle A can be given by vector sum of force experienced by this particle due to all the other q charges on vertices of the cube. For this we use vector form of coulomb's law F r1 r2 kq1 2 r1 r2 3 kq2 akˆ From the figure the different forces acting on A are given as FA1 = a 3 kq2 Z FA2 aˆj akˆ kq2 aˆi aˆj akˆ kq2 aˆi akˆ (a,0,a) 1 , FA3 ; FA4 3 3 3 4 (a,a,a) (0,a,a) 2a 3a 2a 2 3 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 kq2 aˆi aˆj X5 A (0,0,0) , FA6 (a,0,0) FA5 3 (0,a,0) kq2 aˆi kq2 aˆj 6 7 a3 2a , FA7 a3 (a,a,0) y The net force experienced by A can be given as kq2 1 1 1 3 3 2 Fnet FA1 FA2 FA3 FA4 FA5 FA6 FA7 a 2 ˆi ˆj kˆ E7
JEE-Physics Example Five point charges, each of value +q are placed on five vertices of a regular hexagon of side Lm. What is the magnitude of the force on a point charge of value –q coulomb placed at the centre of the hexagon? Solution If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on –q at O will be zero (as the forces due to individual charges will balance each other). Now if f is the force due to sixth charge and F due to remaining five charges. 1 qq q2 F f 0 F f F f 4 0 L2 4 0 L2 ELECTRIC FIELD In order to explain ‘action at a distance’, i.e., ‘force without contact’ between charges it is assumed that a charge or charge distribution produces a field in space surrounding it. So the region surrounding a charge (or charge distribution) in which its electrical effects are perceptible is called the electric field of the given charge. Electric field at a point is characterized either by a vector function of position E called ‘electric intensity’ or by a scalar function of position V called ‘electric potential’. The electric field in a certain space is also visualized graphically in terms of ‘lines of force.’ So electric intensity, potential and lines of force are different ways of describing the same field. Intensity of electric field due to point charge p Electric field intensity is defined as force on unit test charge. r q Lim F = kq ˆr = kq E q0 r2 r3 r q0 0 Note : Test charge (q ) is a fictitious charge that exerts no force on nearby charges but experiences forces due to them. 0 Properties of electric field intensity : NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 (i) It is a vector quantity. Its direction is the same as the force experienced by positive charge. (ii) Electric field due to positive charge is always away from it while due to negative charge always towards it. (iii) Its unit is Newton/coulomb (iv) Its dimensional formula is [MLT–3A–1] (v) Force on a point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge. E qE (vi) It obeys the superposition principle that is the field intensity point due to charge distribution is vector sum of the field intensities due to individual charge 8E
JEE-Physics GOLDEN KEY POINTS • Charged particle in an electric field always experiences a force either it is at rest or in motion. 1 • In presence of a dielectric , electric field decreases and becomes r times of its value in free space. • Test charge is always a unit (+ ve) charge. E Ftest test charge • If identical charges are placed on each vertices of a regular polygon, then E at centre = zero. ELECTRIC FIELD INTENSITIES DUE TO VARIOUS CHARGE DISTRIBUTIONS Due to discrete distribution of charge p Field produced by a charge distribution for discrete distribution:– kq q4 ri By principle of superposition intensity of electric field due to ith charge E ip ri3 ri q3 qi q1 q2 Net electric field due to whole distribution of charge Ep Ei P r i 1 Continuous distribution of charge 1 dq Treating a small element as particle E 4 0 r r3 dq Due to linear charge distribution E = k d r2 [ = charge per unit length] Due to surface charge distribution E = k ds s r2 [ = charge per unit area] Due to volume charge distribution E = k dv v r2 [ = charge per unit volume] NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Electric field strength at a general point due to a uniformly charged rod P As shown in figure, if P is any general point in the surrounding of rod, 1 2 to find electric field strength at P, we consider an element on rod of r length dx at a distance x from point O as shown in figure. Now if dE be the electric field at P due to the element, then kdq Q O dE = Here dq = dx L x2 r2 L dEcos dE Electric field strength in x–direction due to dq at P is kdq kQ sin P dEsin sin dx dE r x = dE = x2 r2 sin = L x2 r2 Here we have x = r tan and dx = r sec2 d Thus dE kQ r sec2 d kQ x sin , Strength = L r dx x L r2 sec2 sin d E9
JEE-Physics Net electric field strength due to dq at point P in x–direction is kQ 1 sin d kQ cos 1 kQ 2 Lr 2 Lr Lr Ex dE x cos 2 cos 1 kQ dx Similarly, electric field strength at point P due to dq in y–direction is dE y dE cos = × cos L r2 x2 Again we have x = r tan and dx = r sec2 d. Thus we have dE = kQ cos × r sec2 d = kQ cos d y L r2 sec2 Lr Net electric field strength at P due to dq in y–direction is 1 kQ kQ cos d = k Q Lr Lr 2 L r 1 Ey = 2 dE y = sin = sin 1 sin 2 Thus electric field at a general point in the surrounding of a uniformly charged rod which subtend angles 1 and 2 at the two corners of rod can be given as in x–direction : Ex = kQ cos 2 cos 1 and in y–direction E y kQ sin 1 sin 2 Lr Lr Electric field due to a uniformly charged ring Case – I : At its centre B + + + A+ + + + ++++ + ++ + + O R + + + D+ NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 C + + + ++ Here by symmetry we can say that electric field strength at centre due to every small segment on ring is+ cancelled by the electric field at centre due to the segment exactly opposite to it. The electric field strength at centre due to segment AB is cancelled by that due to segment CD. This net electric field strength at the centre of a uniformly charged ring is E =0 0 Case II : At a point on the axis of Ring d + + + ++ + + + +++++++ R P dE cos C + dE sin dE Q E 10
JEE-Physics Here we'll find the electric field strength at point P due to the ring which is situated at a distance x from the ring centre. For this we consider a small section of length d on ring as shown. The charge on this elemental section is dq = Q d [Q= total charge of ring] 2 R Kdq Due to the element d, electric field strength dE at point P can be given as dE= R2 x2 The component of this field strength dE sin which is normal to the axis of ring will be cancelled out due to the ring section opposite to d . The component of electric field strength along the axis of ring dE cos due to all the sections will be added up. Hence total electric field strength at point P due to the ring is 2R kdq x 2R kQx d = kQx 2 R R2 x2 = O 2R R2 x2 2 R R2 x2 3 / 2 d O E p dE cos = R2 x2 × O kQ x kQ x R2 x2 R2 x2 3 /2 = 3/2 2R 2 R Electric field strength due to a charged circular arc at its centre : Figure shows a circular arc of radius R which subtend an angle at its centre. To find electric field strength at C, we consider a polar segment on arc of angular width d at an angle from the angular bisector XY as shown. x Rd +++++++++++++++++++ d NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65+++ ++ +++++ R C dEsin dE dEcos Y The length of elemental segment is Rd, the charge on this element d is dq = Q .d Due to this dq, electric field at centre of arc C is given as dE kdq R2 Now electric field component due to this segment dEsin which is perpendicular to the angular bisector gets cancelled out in integration and net electric field at centre will be along angular bisector which can be calculated by integrating dEcos within limits from – 2 to 2 . Hence net electric field strength at centre C is EC dE cos / 2 kQ kQ / 2 kQ / 2 kQ sin 2 kQ s in / 2 R 2 R 2 R 2 / 2 R 2 2 2 2 cos d = / 2 = cos d = sin = sin = R 2 E 11
JEE-Physics Electric field strength due to a uniformly surface charged disc : If there is a disc of radius R, charged on its surface with surface charge density we wish to find electric field strength due to this disc at a distance x from the centre of disc on its axis at point P shown in figure. dy yx dE P To find electric field at point P due to this disc, we consider an elemental ring of radius y and width dy in the disc as shown in figure. The charge on this elemental ring dq = .2ydy [Area of elemental ring ds = 2y dy] Now we know that electric field strength due to a ring of radius R, charge Q, at a distance x from its centre on kQx x2 R2 3 /2 its axis can be given as E Here due to the elemental ring electric field strength dE at point P can be given as kdqx k2y dy x dE= x 2 y 2 3 / 2 = x2 y2 3 / 2 Net electric field at point P due to this disc is given by integrating above expression from 0 to R as R k 2 x y dy R 2 y dy = 2kx 1 R 1 x x2 y2 3 /2 x2 2 0 dE 0 0 E = = 3/2 kx y2 x2 R2 x2 y2 0 Example Calculate the electric field at origin due to infinite number of charges as shown in figures below. fig (a) fig (b) qq q q –q q O 4 x(m) 4 x(m) O 12 12 Solution E0 kq 1 1 1 kq.1 4kq [ S = a 1 (a) 1 4 = (1 1 / 4) = 3 , a = 1 and r = ] 1r 4 16 (b) E0 kq 1 1 1 kq.1 4kq NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 1 4 = 5 1 1 / 4 = 16 Example A charged particle is kept in equilibrium in the electric field between the plates of millikan oil drop experiment. If the direction of the electric field between the plates is reversed, then calculate acceleration of the charged particle. Solution Let mass of the particle = m, Charge on particle = q Intensity of electric field in between plates = E, Initially mg = qE After reversing the field ma = mg + qE ma = 2mg Acceleration of particle a = 2g 12 E
JEE-Physics Example Calculate the electric field intensity E which would be just sufficient to balance the weight of an electron. If this electric field is produced by a second electron located below the first one what would be the distance between them? [Given : e = 1.6 × 10-19 C, m = 9.1 × 10–31 kg and g = 9.8 m/s2] Solution As force on a charge e in an electric field E F = eE e So according to given problem F = W eE = mg e mg 9.1 1031 9.8 V E e 1.6 1019 = 5.57 × 10–11 m As this intensity E is produced by another electron B, located at a distance r below A. 1e e So, r 9 109 1.6 1019 5 m E 40 r2 r 4 0 E 5.57 10 11 Example A block having mass m = 4 kg and charge q = 50 C is connected to a spring having a force constant k = 100 N/m. The block lies on a frictionless horizontal track and a uniform electric field E = 5 × 105 V/m acts on the system as shwon in figure. The block is released from rest when the spring is unstretched (at x = 0) (a) By what maximum amount does the spring expand? (b) What is the equilibrium position of the block? (c) Show that the block's motion is simple harmonic and determine the amplitude and time period of the motion. Solution (a) As x increases, electric force qE will accelerate the block while elastic force in the spring kx will oppose the motion. The block will move away from its initial position x = 0 till it comes to rest, i.e., work done by the electric force is equal to the energy stored in the spring. So if x is maximum stretch of the spring. max 1 k x 2 (qE )xmax xmax = 2qE x = 2 (50 106 ) (5 105 ) = 0.5 m 2 m k 100 ax max (b) In equilibrium position F = 0, so if x is the stretch of the spring in equilibrium position R0 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 k x = qE x = ( qE/k ) = 1 x = 0.25 m max 0 0 2 (c) If the displacement of the block from equilibrium position (x ) is x, restoring force will be 0 F = k(x ± x ) qE = kx [ax kx = qE] 0 0 and as the restoring force is linear the motion will be simple harmonic with time period m4 T 2 2 0.4 s k 100 and amplitude = x – x = 0.5 – 0.25 = 0.25 m max 0 E 13
JEE-Physics ELECTRIC LINES OF FORCE Electric lines of electrostatic field have following properties + A B– (i) Imaginary qA>qB (ii) Can never cross each other (iii) Can never be closed loops (iv) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system (1/o) electric lines are associated with unit charge, so if a body encloses a charge q , total lines of force associated with it (called flux) will be q/o. (v) Lines of force ends or starts normally at the surface of a conductor. (vi) If there is no electric field there will be no lines of force. (vii) Lines of force per unit area normal to the area at a point represents magnitude of intensity , crowded lines represent strong field while distant lines weak field. (viii) Tangent to the line of force at a point in an electric field gives the direction of intensity. So a positive charge free to move follow the line of force. GOLDEN KEY POINTS • Lines of force starts from (+ve) charge and ends on (–ve) charge. • Lines of force start and end normally on the surface of a conductor . +s + – –s ++ – E=0 E=0 + – – – – – – – + – + – fixed point charge near edge effect infinite metal plate • The lines of force never intersect each other due to superposition principle. • The property that electric lines of force contract longitudinally leads to explain attraction between opposite charges. • The property that electric lines of force exert lateral pressure on each other leads to explain repulsion between like charges. Electric flux () NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 The word \"flux\" comes from a Latin word meaning \"to flow\" and you can consider the flux of a vector field to be a measure of the flow through an imaginary fixed element of surface in the field. Electric flux is defined as E E dA This surface integral indicates that the surface in question is to be divided into infinitesimal elements of area dA and the scalar quantity E dA is to be evaluated for each element and summed over the entire surface. Important points about electric flux : (i) It is a scalar quantity (ii) Units (V–m) and N – m2/C Dimensions : [ML3T–3A–1] (iii) The value of does not depend upon the distribution of charges and the distance between them inside the closed surface. 14 E
JEE-Physics Electric Flux through a circular Disc : Figure shows a point charge q placed at a distance from a disc of radius R. Here we wish to find the electric flux through the disc surface due to the point charge q. We know a point charge q originates electric flux in radially outward direction. The flux is originated in cone shown in figure passes through the disc surface. q To calculate this flux, we consider on elemental ring an disc surface of radius x and width dx as shown. Area of kq this ring (strip) is dS 2x dx . The electric field due to q at this elemental ring is given as E= x 2 2 If d is the flux passing through this elemental ring, then dS x q kq 2kq x dx d = E d S c o s = x2 2 × 2x dx × = 2 x2 3 / 2 x2 2 R q R q 1 R q 1 1 = 0 = 2 0 2 2 0 d = q x dx x dx = x2 3 / 2 0 2 x2 0 2 0 x2 2 x2 3 / 2 2 0 2 The above result can be obtained in a much simpler way by using the concept of solid angle and Gauss's law. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Electric flux through the lateral surface of a cylinder due to a point charge : Figure shows a cylindrical surface of length L and radius R. On its axis at its centre a point charge q is placed. Here we wish to find the flux coming out from the lateral surface of this cylinder due to the point charge q. For this we co ns id er an elem ental s tr ip o f wi dth d x on the s ur face of c ylinder as s hown. T he area of th is s tr ip i s dS = 2 R .d x dS E C x R q L E 15
JEE-Physics kq The electric field due to the point charge on the strip can be given as E x2 R2 . If d is the electric flux R dx x2 R2 = 2 KqR2 × x2 R 2 3 / 2 Kq through the strip, then d EdS cos = x2 R2 × 2 Rdx × qR2 L / 2 dx q 0 d L / 2 x2 R2 2 4R2 2 0 Total flux through the lateral surface of cylinder 3/2 = This situation can also be easily handled by using the concepts of Gauss's law. GAUSS'S LAW It relates with the total flux of an electric field through a closed surface to the net charge enclosed by that surface and according to it, the total flux linked with a closed surface is 1 / 0 times the charge enclosed by the closed q surface i.e., S E.ds = 0 n E ds r rO +q REGARDING GAUSS'S LAW IT IS WORTH NOTING THAT : Note : (i) Flux through gaussian suface is independent of its shape. (ii) Flux through gaussian suface depends only on charges present inside gaussian surface. (iii) Flux through gaussian suface is independent of position of charges inside gaussian surface. (iv) Electric field intensity at the gaussian surface is due to all the charges present (inside as well as out side) (v) In a close surface incoming flux is taken negative while outgoing flux is taken positive. (vi) In a gaussian surface = 0 does not employ E = 0 but E = 0 employs = 0. (vii) Gauss's law and Coulomb's law are equivalent, i.e., if we assume Coulomb's law we can prove Gauss's law and NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 vice–versa. To prove Gauss's law from Coulomb's law consider a hypothetical spherical surface [called Gaussian– surface] of radius r with point charge q at its centre as shown in figure. By Coulomb's law intensity at a point P on the surface will be, 1 And hence electric flux linked with area = 1 q E 4 0r3 r ds E.ds 4 0 r3 r.ds 1 q 1 1 q r ds E.ds 4 0 r2 4 0 r2 E.ds 0 Here direction of S and are same, i.e., S = ds 4 r 2 S = Which is the required result. Though here in proving it we have assumed the surface to be spherical, it is true for any arbitrary surface provided the surface is closed. 16 E
JEE-Physics (viii) (a) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non–uniform) total flux linked with it will be zero. sphere E=0 E=0 (A) (B) (b) If a closed body encloses a charge q, then total flux q q (C) q q (A) (B) E=(q/0) (D) E=(q/ 0) E=(q/ 0) E=(q/0) linked with the body will be S = q E.ds 0 From this expression it is clear that the flux linked with a closed body is independent of the shape and size of the body and position of charge inside it.[figure] Note : So in case of closed symmetrical body with charge at its centre, flux linked with each half will be 1 E = q and the symmetrical closed body has n identical faces with point charge at its centre, flux linked 2 20 with each face will be E = q n n0 (ix) Gauss's law is a powerful tool for calculating electric intensity in case of symmetrical charge distribution by choosing a Gaussian– surface in such a way that E is either parallel or perpendicular to its various faces. As an example, consider the case of a plane sheet of charge having charge density . To calculate E at a point P close to it consider a Gaussian surface in the form of a 'pill box' of cross–section S as shown in figure. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 +++ n E + + + ++ n E + + + ++ + + E n + + + ++ + + n + + + ++ + + + + + ++ + + Ein=0 + + + ++ + + n+ + + + + + + + + + ++ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + + + + +++ Sheet of charge Conductor (A) (B) The charge enclosed by the Gaussian–surface = S and the flux linked with the pill box = ES + 0 + ES = 2ES (as E is parallel to curved surface and perpendicular to plane faces) 11 So from Gauss's law, E = 0 (q), 2ES = 0 S E = 20 E 17
JEE-Physics (x) If =0, =0, so q = 0 but if q=0, =0 So may or may not be zero. E E E.ds E.ds If a dipole is enclosed by a closed surface then, q =0, so =0, but 0 E.ds E Note : If instead of plane sheet of charge, we have a charged conductor, then as shown in figure (B) E =0. in So E ES and hence in this case E= 0 . This result can be verified from the fact that intensity at the surface 1q of a charged spherical conductor of radius R is, E = 40 R2 with q = 4R2 1 = 0 So for a point close to the surface of conductor, E = 40R2 × 4 R2 Example If a point charge q is placed at the centre of a cube. What is the flux linked (a) with the cube? (b) with each face of the cube? Solution (a) According to Gauss's law flux linked with a closed body is (1/0) times the charge enclosed and here the 1 closed body cube is enclosing a charge q so, T 0 (q) (b) Now as cube is a symmetrical body with 6–faces and the point charge is at its centre, so electric flux linked with each face will be F 1 T q 6 60 Note: (i) Here flux linked with cube or one of its faces is independent of the side of cube. (ii) If charge is not at the centre of cube (but anywhere inside it), total flux will not change, but the flux linked with different faces will be different. Example NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 If a point charge q is placed at one corner of a cube, what is the flux linked with the cube? Solution In this case by placing three cubes at three sides of given cube and four cubes above, the charge will be in the centre. qq So, the flux linked with each cube will be one–eight of the flux 0 . Flux associated with given cube = 80 18 E
JEE-Physics ds E FLUX CALCULATION USING GAUSS LAW R E ds ds in= –R2 E and out= R2 E total = 0 E RE in= circular = –R2 E and out= curved = R2 E total = 0 E y E ds ds a in= –a2 E and out= a2 E total = 0 a ds x z a E in 1 R 2E and out 1 R 2E total = 0 – 2 2 E kq 2R 2 q R2 q Rq E R2 0 2 0 4 Note : here electric field is radial q q q = 2 0 hemisphere q qq = 2 0 cylinder R R q q q cube = 2 0 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 q q qq = q q 8 0 E q q = 4 0 19
JEE-Physics Example As shown in figure a closed surface intersects a spherical conductor. If a negative charge is placed at point P. What is the nature of the electric flux coming out of the closed surface ? close surface conductor ++++ ++++++++ P –Q Solution + Point charge Q induces charge on conductor as shown in figure. + Net charge enclosed by closed surface is negative so flux is negative. +++ Example Consider E = 3 × 103 ˆi (N/C) then what is the flux through the square of 10 cm side, if the normal of its plane makes 60° angle with the X axis. Solution = EScos = 3 × 103 × [10 × 10–2]2 × cos60° = 3 × 103 × 10–2 × ½ = 15 Nm2/C Example Find the electric field due to an infinitely long cylindrical charge distribution of radius R and having linear charge density at a distance half of the radius from its axis. Solution F I2kr 2k R R HG KJE= R2 = R2 2 = 4 0 R r = point will be inside so 2 ELECTRIC FIELD DUE TO SOLID CONDUCTING OR HOLLOW SPHERE ++ E • For outside point (r > R) P dA r +++ +++ q OR E.ds Using Gauss's theorem 0 ++ At every point on the Gaussian surface ; = E ds cos 0° = E ds Gaussian surface NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 E ds E.ds E.ds q [E is constant over the gaussian surface] E 4r2 q E = q r2 p 40 0 0 • For surface point r = R : q • For Inside point (r < R) : ES 4 0 R2 Because charge inside the conducting sphere or hollow is zero. (i.e. q = 0) So q = 0 E = 0 E .ds 0 in 20 E
JEE-Physics ELECTRIC FIELD DUE TO SOLID NON CONDUCTING SPHERE E P ds • Outside (r > R) r From Gauss's theorem OR s q E 4r2 q EP q r2 Gaussian surface E.ds 0 E 0 4 0 P ds r • At surface (r = R) OR q ES 40 R2 Put r = R • Inside (r < R) : From Gauss's theorem q E.ds 0 s Where q charge contained within Gaussian surface of radius r E(4r2 ) q E q ...(i) Gaussian surface 0 4r2 0 As the sphere is uniformly charged, the volume charge density (charge/volume) is constant throughout the q 4 r 3 q 4 r3 qr3 3 3)R3 3 q sphere 4 R 3 charge enclosed in gaussian surface q = (4 R3 3 1 qr 4 0 R3 put this value in equation (i) E in ELECTRIC FIELD DUE TO AN INFINITE LINE DISTRIBUTION OF CHARGE Let a wire of infinite length is uniformly charged having a constant linear charge density . P is the point where electric field is to be calculated. Let us draw a coaxial Gaussian cylindrical surfaces of length . dS1 From Gauss's theorem q r Gaussian E.dS1 E surface s1 dS3 E.dS2 E.dS 3 0 E s2 s3 E dS1 so E.dS1 0 and E dS2 so E.dS2 0 E 2r q E 0 [ E dS3 ] dS2 Charge enclosed in the Gaussian surface q = . NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 So E 2r E r or E 2k where k = 1 0 2 0 r 4 0 ( f ) Charged cylindrical nonconductor of infinite length Electric field at outside point 2k ˆr r > R Gaussian Electric field at inside point EA r r < R Surface r EB 2 0 R2 E 21
JEE-Physics DIELECTRIC IN ELECTRIC FIELD . Let E0 be the applied field, Due to polarisation, electric field is – + E p E0 – E0 + E0 – Ep + The resultant field is E . For homogeneous and isotropic dielectric, – E+ + the direction of E p is opposite to the direction of E0 . – So, Resultant field is E=E0–EP GOLDEN KEY POINTS • Electric field inside a solid conductor is always zero. • Electric field inside a hollow conductor may or may not be zero (E 0 if non zero charge is inside the sphere). • The electric field due to a circular loop of charge and a point charge are identical provided the distance of the observation point from the circular loop is quite large as compared to its radius i.e. x >>> R. Example For infinite line distribution of charge draw the curve between log E and log r. Solution E r A where A constant log E 2 0 r 2 0 log A take log on both side log E = log A – log r log r Example A point charge of 0.009 C is placed at origin. Calculate intensity of electric field due to this point charge at point 2, 7,0 . Solution ; where x i yj 2 i 7 j , 9 105 9 109 ( 2 ˆi 7 ˆj) = 3 2 ˆi 3 7 ˆj NC 1 E qr r E (3)3 4 0 r3 ELECTROSTATIC POTENTIAL ENERGY NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Potential energy of a system of particles is defined only in conservative fields. As electric field is also conservative, we define potential energy in it. Potential energy of a system of particles we define as the work done in assembling the system in a given configuration against the interaction forces of particles. Electrostatic potential energy is defined in two ways. (i) Interaction energy of charged particles of a system (ii) Self energy of a charged object • Electrostatic Interaction Energy Electrostatic interaction energy of a system of charged particles is defined as the external work required to assemble the particles from infinity to the given configuration. When some charged particles are at infinite separation, their potential energy is taken zero as no interaction is there between them. When these charges are brought close to a given configuration, external work is required if the force between these particles is repulsive and energy is supplied to the system, hence final potential energy of system will be positive. If the force between the particle is attractive, work will be done by the system and final potential energy of system will be negative. 22 E
JEE-Physics • Interaction Energy of a system of two charged particles Figure shows two + ve charges q and q separated by a distance r. The electrostatic interaction energy of this 12 system can be given as work done in bringing q from infinity to the given separation from q . 21 x q1 r q2 dx F r r kq1q2 F.dx x2 It can be calculated as W = dx [ – ve sign shows that x is decreasing] = – W = kq1q2 = U [ interaction energy] r If the two charges here are of opposite sign, the potential energy will be negative as U = – kq1q2 r • Interaction Energy for a system of charged particles When more than two charged particles are there in a system, the interaction energy can be given by sum of interaction energies of all the pairs of particles. For example if a system of three particles having charges q , 1 q and q is given as shown in figure. 23 q1 r3 r2 r1 q2 q3 The total interaction energy of this system can be given as U = kq1q2 + kq1q3 + kq2q3 r3 r2 r1 ELECTRIC POTENTIAL Electric potential is a scalar property of every point in the region of electric field. At a point in electric field potential is defined as the interaction energy of a unit positive charge. If at a point in electric field a charge q 0 U has potential energy U, then electric potential at that point can be given as V = q0 joule/coulomb NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Potential energy of a charge in electric field is defined as work done in bringing the charge from infinity to the given point in electric field. Similarly we can define electric potential as \"work done in bringing a unit positive charge from infinity to the given point against the electric forces. Example A charge 2C is taken from infinity to a point in an electric field, without changing its velocity, if work done against electrostatic forces is –40J then potential at that point is ? Solution V Wext 40J 20V q 2C Note : Always remember to put sign of W and q. E 23
JEE-Physics • Electric Potential due to a point charge in its surrounding : p r q U The potential at a point P at a distance r from the charge q V = q0 . Where U is the potential energy of P charge q at point p, U = kqq0 . Thus potential at point P is kq 0r V = Pr Electric Potential due to a charge Rod : Figure shows a rod of length L, uniformly charged with a charge Q. Due to this we'll find electric potential at a point P at a distance r from one end of the rod as shown in figure. x Q dx rP L For this we consider an element of width dx at a distance x from the point P. Charge on this element is Q dQ = L dx The potential dV due to this element at point P can be given by using the result of a point charge as kdq kQ dV = = dx x Lx rL kQ kQ r L Net electric potential at point P : V = dV = Lx dx = L ln r r Electric potential due to a charged ring Case – I : At its centre To find potential at the centre C of the ring, we first find potential dV at centre due to an elemental charge kdq dV = kdq = kQ dq on ring which is given as dV = Total potential at C is V= R . R R Q ++ ++ R++ ++ ++ ++ dq C NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 + + + ++ As all dq's of the ring are situated at same distance R from the ring centre C, simply the potential due to all dq's kQ is added as being a scalar quantity, we can directly say that the total electric potential at ring centre is . R Here we can also state that even if charge Q is non–uniformly distributed on ring, the electric potential C will remain same. 24 E
JEE-Physics Case II : At a point on axis of ring We find the electric potential at a point P on the axis of ring as shown, we can directly state the result as here also all points of ring are at same distance x2 R2 from the point P, thus the potential at P can be given kQ as V = R2 x2 P ++++++ ++ + + + + x 2+R 2 R xP + Electric potential due to a uniformly charged disc : Figure shows a uniformly disc of radius R with surface charge density coul/m2. To find electric potential at point P we consider an elemental ring of radius y and width dy, charge on this elemental ring is dq=2y dy. kdq k..2y dy Due to this ring, the electric potential at point P can be given as dV = x2 y2 = x2 y2 dy yx P Net electric potential at Point P due to whole disc can be given as R y dy R dV 0 2 0 · x2 y2 = 2 0 2 0 V = x2 y2 = x2 R2 x 0 ELECTRIC POTENTIAL DUE TO HOLLOW OR CONDUCTING SPHERE • At outside sphere NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 R r>R P According to definition of electric potential, electric potential at point P r rq q q r 1 q 1 r q E .dr 4 0 r2 4 0 r 4 0 4 0 4 0 V r2 dr Eout r2 ; V dr r E 25
JEE-Physics • At surface R Rq q q R 1 q 1 R q E .dr 4 0 r2 4 0 r 4 0 4 0 4 0 V r2 dr E out ; V r2 dr V R • Inside the surface : Inside the surface E dV 0 V = constant so V q dr 4 0 R ELECTRIC POTENTIAL DUE TO SOLID NON CONDUCTING SPHERE • At outside sphereSame as conducting sphere. • At Surface Same as conducting sphere. • Inside the sphere r R r V E .dr V E1dr E 2 dr R R kq r kqr dr 1 R kq r2 r V r2 dr V kq r R3 2 R 3 R R V 1 r2 R2 V kq 3 R 2 r2 kq 2R3 R 2R3 2R3 Potential Difference Between Two points in electric field Potential difference between two points in electric field can be defined as work done in displacing a unit positive charge from one point to another against the electric forces. AB If a unit +ve charge is displaced fromV Aa point A to B as shown work reVqBuired can be given as V – V = – B B A E.dx A If a charge q is shifted from point A to B, work done against electric forces can be given as W = q (V – V ) BA If in a situation work done by electric forces is asked, we use W = q (V – V ) AB If V < V , then charges must have tendency to move toward B (low potential point) it implies that electric BA forces carry the charge from high potential to low potential points. Hence we can say that in the direction of electric field always electric potential decreases. Example 1C charge is shifted from A to B and it is found that work done by external force is 80J against NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 electrostic forces, find V – V AB Solution W = q(V – V ) AB B A 80 J = 1C (V – V ) V – V = – 80 V B A A B Equipotential surfaces For a given charge distribution , locus of all points having same potential is called 'equipotential surface'. • Equipotential surfaces can never cross each other (otherwise potential at a point will have two values which is absurd) 26 E
JEE-Physics • Equipotential surfaces are always perpendicular to direction of electric field. • If a charge is moved from one point to the other over an equipotential surface then work done W = – U = q (V –V ) = 0 [ V = V ] AB AB BA BA • Shapes of equipotential surfaces V=V1 V=V2 V=V1 V=V2 V=V2 V=V1 for uniform electric field for a point charge, spherical for a line distribution of equipotential surfaces conductor equipotential charge equipotential surfaces are spherical are parallel plane surfaces are cylinderical • The intensity of electric field along an equipotential surface is always zero. Electric Potential Gradient The maximum rate of change of potential at right angles to an equipotential surface in an electric field is defined as potential gradient. E V = – grad V Note : Potential is a scalar quantity but the gradient of potential is a vector quantity L O MN QPIn cartesian co–ordinates V = V i V j V k x y z Example If V = –5x + 3y + 15 z then find magnitude of electric field at point (x,y,z). Solution – V ˆi V ˆj V kˆ = –(–5 i + 3 j + 15 k ) 25 9 15 49 7 unit E x y z |E| Example The four charges q each are placed at the corners of a square of side a. Find the potential energy of one of the charges NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 The electric potential of a point A due to charges B, C and D is 1q 1 q 1 q 1 2 1 q 2a 40 a 40 2 a V = 40 a + 40 + = Potential energy of the charge at A is PE = qV = 1 2 1 q2 . 40 2 a E 27
JEE-Physics Example A proton moves with a speed of 7.45 × 105 m/s directly towards a free proton originally at rest. Find the distance of closest approach for the two protons. 1 N m2 ; m = 1.67 × 10–27 kg and e = 1.6 × 10–19 C Given : 4 0 = 9 × 109 p C2 Solution As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving and so the velocity of first particle will decrease while of other will increase and at closect approach both will move with same velocity. So if v is the common velocity of each particle at closest approach, by 'conservation of momentum'. 1 mu = mv + mv v = 2 u 1 1 1 1 e2 And by conservation of energy 2 mu2 = 2 mv2 + 2 + 40 r 4e2 u So, r= 40mu2 [as v = 2 ] And hence substituting the given data, 4 (1.6 10 19 )2 r = 9 × 109 × 1.67 1027 (7.45 103 )2 = 10–12 m ELECTRIC DIPOLE A system of two equal and opposite charges separated by a certain distance is called electric dipole, shown in figure. Every dipole has a characteristic property called dipole moment. It is defined as the product of magnitude of either charge and the separation between the charges, given as p = qd d -q p +q In some molecules, the centres of positive and negative charges do not coincide. This results in the formation of electric dipole. Atom is non – polar because in it the centres of positive and negative charges coincide. Polarity can be induced in an atom by the application of electric field. Hence it can be called as induced dipole. • Dipole Moment : Dipole moment p = q d +q p q NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 (i) Vector quantity , directed from negative to positive charge (ii) Dimension : [LTA], Units : coulomb × metre (or C-m) (iii) Practical unit is \"debye\" Two equal and opposite point charges each having charge 10–10 frankline ( e) and separation of 1Å then the value of dipole moment (p ) is 1 debye. Cm E 1 Debye = 10–10 × 10–10 Fr × m = 10–20 × 3 109 3.3 × 10–30 C × m 28
JEE-Physics Example A system has two charges qA = 2.5 × 10–7 C and qB = – 2.5 × 10–7 C located at points A: (0, 0, – 0.15 m) and B ; (0, 0, + 0.15 m) respectively. What is the total charge and electric dipole moment of the system? Solution Total charge = 2.5 × 10–7 – 2.5 × 10–7 = 0 Electric diople moment, p = Mangitude of either charge × separation between charges = 2.5 × 10–7 [0.15 + 0.15] C m = 7.5 × 10–8 C m. The direction of dipole moment is from B to A. • Dipole Placed in uniform Electric Field Figure shows a dipole of dipole moment p placed at an angle to the direction of electric field. Here the charges of dipole experience forces qE in opposite direction as shown. Fnet = qE (q) E 0 + qE p E – – qE Thus we can state that when a dipole is placed in a uniform electric field, net force on the dipole is zero. But as equal and opposite forces act with a separation in their line of action, they produce a couple which tend to align the dipole along the direction of electric field. The torque due to this couple can be given as = Force × separation between lines of actions of forces = qE × d sin = pE sin r F d qE qd E p E Work done in Rotation of a Dipole in Electric field When a dipole is placed in an electric field at an angle , the torque on it due to electric field is pE sin Work done in rotating an electric dipole from 1 to 2 [ uniform field] 2 dW = d so W = dW = d and W1 2 = W = pE sin d = pE (cos1 – cos2) 1 e.g. W = pE [1– (–1)] = 2 pE W = pE (1–0) = pE 0 180 0 90 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 If a dipole is rotated from field direction ( = 0° ) to then W = pE ( 1– cos) E + –+ E – p+ p p = 0 E = 180° = minimum = 0 = minimum = 0 W = minimum = 0 – W = maximum = 2pE = 90° = maximum = pE W = pE E 29
JEE-Physics Electrostatic potential energy : Electrostatic potential energy of a dipole placed in a uniform field is defined as work done in rotating a dipole from a direction perpendicular to the field to the given direction i.e., W90 pE sin d = – pE cos = – p.E 90 E is a conservative field so what ever work is done in rotating a dipole from 1 to 2 is just equal to change in electrostatic potential energy W1 2 U 2 U 1 = pE (cos 1– cos 2) Work done in rotating an electric dipole in an electric field Suppose at any instant, the dipole makes an angle with the electric field. The torque acting on dipole. = qEd = (q 2 sin)E = pE sin The work done in rotating dipole from 1 to 2 2 2 B +q W d pE sin d qE 1 1 2 d W = pE (cos1 – cos2) = U – U ( U = – pE cos) 2 1 q A C E qE Force on an electric dipole in Non–uniform electric field : If in a non–uniform electric field dipole is placed at a point where electric field is E, the interaction energy of dipole at this point U . Now the force on dipole due to electric field F = – U p.E r dE If dipole is placed in the direction of electric field then F= – p dx Example Calculate force on a dipole in the surrounding of a long charged wire as shown in the figure. +++++++++++ p -q +q NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 r Solution In the situation shown in figure, the electric field strength due to the wire, at the position 2k of dipole as E = r Thus force on dipole is F = – p. dE = – p 2k = 2kp dr r2 r2 Here –ve charge of dipole is close to wire hence net force an dipole due to wire will be attractive. 30 E
JEE-Physics ELECTRIC POTENTIAL DUE TO DIPOLE • At axial point kq –q O q V1 (r ) P Electric potential due to +q charge r Electric potential due to –q charge kq V2 (r ) kq kq kq 2 kp (r ) (r ) Net electric potential V = V + V = (r2 2 ) r2 2 12 If r > > > V kp r2 • At equitorial point Electric potential of P due to +q charge kq P V = x rx –q O q 1x Electric potential of P due to –q charge kq V = 2x kq kq Net potential V = V + V = = 0 V = 0 12 x x • At general point p cos p . r p q d electric dipole moment V = 40r2 40r3 y Electric field due to an electric dipole Enet Er E Figure shows an electric dipole placed on x–axis at origin. Here we wish to find the electric field and potential at a point O having coordinates (r, ). Due O to the positive charge of dipole electric field at O is in radially outward direction and due to the negative charge it is radially inward as shown in figure. r V 2kp cos 1 V kp sin x r r3 r3 -q +q NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 E =– and E r r Thus net electric field at point O, E = E 2 E 2 = kp 1 3 cos2 net r r3 If the direction of E is at an angle from radial direction, then = tan–1 E 1 tan net Er = 2 Thus the inclination of net electric field at point O is ( E 31
JEE-Physics Oq • At a point on the axis of a dipole : E2 E1 –q r Electric field due to +q charge E = kq Electric field due to –q charge E = kq 1 2 (r )2 (r )2 kq kq kq 4r Net electric field E = E – E = = [ p = q × 2 = Dipole moment] 12 (r )2 (r )2 (r2 2 )2 2kpr 2kp E = (r2 2 )2 If r >>> then E = r3 • At a point on equitorial line of dipole : E1 sin E1 kq kq Electric field due to +q charge E1 x2 ; Electric field due to –q charge E2 x2 E P Vertical component of E and E will cancel each other and horizontal 12 E2 E 2 sin components will be added So net electric field at P E = E1cos + E2cos [ E = E] x x 1 2 E = 2E cos = 2kq cos cos and x r2 2 r 1 x2 x q –q O E 2kq 2kq kp If r > > > then kp kp x3 (r2 2 )3 2 2 )3 2 r3 E r3 = (r2 E or GOLDEN KEY POINTS • For a dipole, potential is zero at equatorial position, while at any finite point E 0 • In a uniform E , dipole may feel a torque but not a force. • If a dipole placed in a field E (Non-Uniform) generated by a point charge, then torque on dipole may be zero, but F 0 • Distribution Point charge Dipole Force between Point charge Dipole and Dipole-dipole Potential proportional to r–1 r–2 point charge E proportional to r–2 r–3 Proportional to r–2 r–3 r–4 Example NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 A short electric dipole is situated at the origin of coordinate axis with its axis along x–axis and equator along y–axis. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal at a point distant r = 5 m from origin. Find the position vector of the point in first quadrant. Solution kp kp cos 1 E P VP r3 1 3 cos2 = r2 1+3 cos2 = 5 cos2 cos = 2 = 450 Position vector 5 r of point P is r = ˆi ˆj 2 32 E
JEE-Physics Example Prove that the frequency of oscillation of an electric dipole of moment p and rotational inertia I for small 1 pE amplitudes about its equilibrium position in a uniform electric field strength E is I 2 Solution Let an electric dipole (charge q and –q at a distance 2a apart) placed in a uniform external electric field of strength E. +q F 2a E P -q F Restoring torque on dipole pE sin = –pE (as is small) PE Here – ve sign shows the restoring tendency of torque. I angular acceleration = I = I pE For SHM 2 comparing we get = I 1 pE Thus frequency of oscillations of dipole n = = I 2 2 ELECTROSTATIC PRESSURE ++++ + ++ ds Force due to electrostatic pressure is directed normally outwards to the surface . ++ Force on small element ds of charged conductor +++ + + ++ 2 dF ds dF = (Charge on ds ) x Electric field = ( ds) 2 0 2 0 Inside E –E = 0 E = E E1 E2 E1 E2 12 1 2 + NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 E2 2 0 Just outside E = E + E = 2E 2 12 (E is field due to point charge on the surface and E is field due to rest of the sphere). 12 The electric force acting per unit area of charged surface is defined as electrostatic pressure. dF 2 Peleectrostatic dS 2 0 E 33
JEE-Physics ++ + + +R+++++ + Equilibrium of liquid charged surfaces (Soap bubble) +++ + + ++ Pressures (forces) act on a charged soap bubble , due to (i) Surface tension P (inward) T (ii) Air outside the bubble P (inward) o (iii) Electrostatic pressure P (outward) e (iv) Air inside the bubble P (outward) i in state of equilibrium inward pressure = outward pressure PT + Po = Pi +Pe Excess pressure of air inside the bubble (P ) = P – P = P – P ex i o T e 4 T 2 4 T 2 4 T 2 but P = and P = P = if P = P then T r e 2 0 ex r 2 0 io r 2 0 Example Brass has a tensile strength 3.5 × 108 N/m2. What charge density on this material will be enough to break it by electrostatic force of repulsion? How many excess electrons per square Å will there then be? What is the value of intensity just out side the surface? Solution dF 2 We know that electrostatic force on a charged conductor is given by ds 20 2 So the conductor will break by this force if, 20 > Breaking strength i.e., 2 2 9 1012 3.5 108 i.e. min 3 7 102 7.94 102 C / m 2 Now as the charge on an electron is 1.6 × 10–19 C, the excess electrons per m2 Further as in case of a conductor near its surface E 7.94 102 8.8 109 V/m 0 9 10 12 CONDUCTOR AND IT'S PROPERTIES [FOR ELECTROSTATIC CONDITION] (i) Conductors are materials which contains large number of free electrons which can move freely inside the c o n d u c to r. (ii) n electrostatics, conductors are always equipotential surfaces. (iii) Charge always resides on outer surface of conductor. (iv) f there is a cavity inside the conductor having no charge then charge will always reside only on outer sur face of conductor. (v) Electric field is always perpendicular to conducting surface. (vi) Electric lines of force never enter into conductors. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 (vii) Electric field intensity near the conducting surface is given by formula E nˆ = 0 A B C nˆ EA 0 nˆ ; E B 0 nˆ and E C 0 (viii) When a conductor is grounded its potential becomes zero. E 34
JEE-Physics (ix) When an isolated conductor is grounded then its charge becomes zero. (x) When two conductors are connected there will be charge flow till their potential becomes equal. 2 (xi) Electric pressure at the surface of a conductor is givey by formula P = 20 where is the local surface charge density. Example Prove that if an isolated (isolated means no charges are near the sheet) large conducting sheet is given a charge then the charge distributes equally on its two surfaces. Solution Let there is x charge on left side of sheet and Q–x charge on right side of sheet. Since point P lies inside the conductor so E = O P x – Q x = 0 2 x Q Q x Q-x x = 2AO 2 A O 2AO = 2AO 2 Q-x P x Q 2A0 2A 0 Q – x = 2 So charge is equally distributed on both sides Example If an isolated infinite sheet contains charge Q on its one surface and charge Q on its other surface then prove 12 that electric field intensity at a point in front of sheet will be Q , where Q = Q + Q 12 2AO Solution = = Q1 nˆ + Q2 nˆ = Q1 Q2 nˆ = Q nˆ EQ1 EQ2 2A0 2A0 2A0 2A0 Electric field at point P : E NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 [This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces]. Example Three large conducting sheets placed parallel to each other at finite distance contains charges Q, –2Q and 3Q respectively. Find electric field at points A, B , C, and D Q -2Q 3Q ·A ·B ·C S o l . E = E + E + E . (i) Here E means electric field due to ‘Q’. A Q –2Q 3Q Q (Q 2Q 3Q) 2Q Q E = 2A0 = 2A0 = A 0 , towards left A E 35
JEE-Physics E = Q (2Q 3Q ) B (ii) 2A0 , towards right = 0 (Q 2Q) (3Q ) 4Q 2Q 2Q 2A0 = A0 , towards right A 0 towards left (iii) E = 2A0 C (Q 2Q 3Q) 2Q Q (iv) E = 2A0 = 2A0 = A0 , towards right D Example Two conducting plates A and B are placed parallel to each other. A is given a charge Q and B a charge Q . 12 Prove that the charges on the inner facing surfaces are of equal magnitude and opposite sign. Also find the charges on inner & outer surfaces. Solution Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore zero. From Gauss’s law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be equal and opposite to that on the inner surface of B. E=0 Q1–q P A Q1 A +q B –q • E Q2+q Q2 B E=0 The distribution should be like the one shown in figure. To find the value of q, consider the field at a point P inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation E = / (20), the electric field at P Q1 q q due to the charge Q – q = 2A0 (downward); due to the charge + q = 2A0 (upward), 1 due to the charge – q = q Q2 q (downward), and due to the charge Q + q = (upward). 2A0 2 2A0 The net electric field at P due to all the four charged surfaces is (in the downward direction) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 E = Q1 q q q Q2 q p 2A0 2A0 2A0 2 A 0 As the point P is inside the conductor, this field should be zero. Hence, Q – q – q + q – Q – q = 0 q Q1 Q2 1 2 2 This result is a special case of the following result. When charged conducting plates are placed parallel to each other, the two outermost, surfaces get equal charges and the facing surfaces get equal and opposite charges. 36 E
JEE-Physics Example Figure shows three large metallic plates with charges – Q, 3Q and Q respectively. Determine the final charges on all the surfaces. –Q 3Q Q Solution We assume that charge on surface 2 is x. Following conservation of charge, we see that surfaces 1 has charge (– Q – x). The electric field inside the metal plate is zero so fields at P is zero. 12 3Q Q -Q-x x 34 56 P Q x x 3Q Q 5Q Resultant field at P : E = 0 2A0 = 2A0 –Q – x = x + 4Q x = 2 P Note : We see that charges on the facing surfaces of the plates are of equal magnitude and opposite sign. This can be in general proved by gauss theorem also. Remember this it is important result. Thus the final charge distribution on all the surfaces is : +3 Q - 5 Q 5 Q +Q - Q +3 Q 2 2 22 22 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Example An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that electric field is perpendicular to sheet and covers all the sheet. Find out charges appearing on its two surfaces. Also Q E E 37
JEE-Physics Solution Let there is x charge on left side of plate and Q – x charge on right side of plate Q-x x Q – x x 2A0 P + E 2A0 x E Qx x Q E x Q EA0 E = 0 2A0 2A0 A0 2A0 2 P QQ So charge on one side is – EA and other side + EA o 2 2 o The resultant electric field on the left and right side of the plate. QQ On right side E = 2 A 0 + E towards right and on left side 2 A 0 – E towards left. SOME OTHER IMPORTANT RESULTS FOR A CLOSED CONDUCTOR. (i) f a charge q is kept in the cavity then –q will be induced on the inner surface and +q will be induced on the outer surface of the conductor (it can be proved using gauss theorem) (ii) If a charge q is kept inside the cavity of a conductor and conductor is given a charge Q then –q charge will be induced on inner surface and total charge on the outer surface will be q + Q. (it can be proved using gauss theorem) +q+Q –q q (iii) Resultant field, due to q (which is inside the cavity) and induced charge on S , at any point outside S (like B,C) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 11 is zero. Resultant field due to q + Q on S and any other charge outside S , at any point inside of surface 22 S (like A, B) is zero 2 .B S2 .C S1 q+Q .q .A – q 38 E
JEE-Physics (iv) Resultant field in a charge free cavity in a closed conductor is zero. There can be charges outside the conduc- tor and on the surface also. Then also this result is true. No charge will be induced on the inner most surface of the conductor. No charge (v). Charge distribution for different types of cavities in conductors S2 S2 S•1Cq •q (A) S1 •C charge is at the common centre (B) (S1, S2 spherical) charge is not at the common centre S2 (S1, S2 spherical) S2 S1 •q S1 • •q C C (C) (D) charge is at the centre of S2 charge is not at the centre of S2 (S2 spherical) (S2 spherical) S2 S2 S1 ·q S1C••q C (F) (E) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 charge is at the centre of charge not at the centre S1 (Spherical) of S1(Spherical) Using the result that Eres in the conducting material should be zero and using result (iii) We can show that Case A BCDEF S1 Uniform Nonuniform Nonuniform Nonuniform Uniform Nonuniform S2 Uniform Uniform Uniform Uniform Nonuniform Nonuniform Note : In all cases charge on inner surface S = – q and on outer surface S = q. The distribution of charge on S will 12 1 not change even if some charges are kept outside the conductor (i.e. outside the surface S ). But the charge 2 distribution on S may change if some charges(s) is/are kept outside the conductor. 2 E 39
JEE-Physics Example An uncharged conductor of inner radius R and outer radius R contains a point charge q at the centre as 12 shown in figure S2 S1 q R1 ·C O R2 ·A ·B (i) Find E and V at points A,B and C (ii) If a point charge Q is kept out side the sphere at a distance ‘r’ (>>R ) from centre then find out 2 resultant force on charge Q and charge q. +q Solution At point A : VA Kq Kq K(q) Kq –q OA R2 R1 , E A OA 3 OA ·q At point B : VB Kq K (q ) Kq Kq , E = 0; At point C : V = Kq Kq OB OB R2 R2 BC OC , E C OC3 OC (ii) Force on point charge Q : KqQ ˆr (r = distance of ‘Q’ from centre ‘O’) FQ r2 Force on point charge q: Fq 0 (using result (iii) & charge on S uniform) 1 Example An uncharged conductor of inner radius R and outer radius R contains a point charge q placed at point P 12 (not at the centre) as shown in figure? Find out the following : B A R1 P S1 C q D R2 S2 (i) V (ii) V (iii) V (iv) E (v) E NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 C A B A B (vi) force on charge Q if it is placed at B Solution Kq K(q) Kq Kq Kq (i) VC CP (ii) VA = R 2 (iii) VB = CB R1 R2 Kq ^ KQq ^ (iv) E = O (point is inside metallic conductor) (v) E = CB (vi) F = CB A B CB2 Q CB2 40 E
JEE-Physics (vi) Sharing of charges : Two conducting hollow spherical shells of radii R and R having charges Q and Q respectively and 1 2 12 seperated by large distance, are joined by a conducting wire. Let final charges on spheres are q and q 12 respectively. q1 q2 R1 R2 Potential on both spherical shell become equal after joining, therefore Kq1 Kq2 ; q1 R1 ...(i) and q + q = Q + Q ......(ii) R1 R2 q2 R2 12 1 2 from (i) and (ii) q1 (Q1 Q2 )R1 ; q2 (Q1 Q2 )R2 R1 R2 R1 R2 q1 R1 ; 1 4 R 2 R1 q2 1 R2 ratio of charges R2 2 4 R 2 2 ratio of surface charge densities 1 R 2 2 R1 Ratio of final charges q1 R1 q2 R2 Ratio of final surface charge densities. 1 R 2 2 R1 Example The two conducting spherical shells are joined by a conducting wire and cut after some time when charge stops flowing.Find out the charge on each sphere after that. Q -3Q R 2R NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 Solution After cutting the wire, the potential of both the shells is equal Kx K(2Q x) kx 2Q Thus, potential of inner shell V = + = in R 2R 2R Kx K(2Q x) KQ –2Q – x and potential of outer shell Vout = 2R + = x 2R R As V = V – KR = Kx – 2Q out in R 2R –2Q = x – 2Q x = 0 So charge on inner spherical shell = 0 and outer spherical shell = – 2Q. E 41
JEE-Physics 5Q 3 -2Q Example Find charge on each spherical shell after joining the inner most shell and 2Q outer most shell by a conducting wire. Also find charges on each surface. 1R 2R 3R Solution 6Q – x Let the charge on the innermost sphere be x. 3 -2Q Finally potential of shell 1 = Potential of shell 3 2x 1R Kx K(2Q) K(6Q x) KQ k –2q k 5Q R 2R 3R 3R 3R 3R 2R 3R 3x –3Q + 6Q – x = 4Q ; 2x = Q ; Q x = Q +3Q/2 2 –3Q/2 –Q/2 Q/2 Q 5Q Charge on innermost shell = , charge on outermost shell = 22 middle shell = –2Q Final charge distribution is as shown in figure. Example 3Q Two conducting hollow spherical shells of radii R and 2R carry charges – -Q R Q and 3Q respectively. How much charge will flow into the earth if inner shell is grounded ? 2R Solution When inner shell is grounded to the Earth then the potential of inner shell will bcome zero because potential of the Earth is taken to be zero. Kx K3Q 3Q + = 0 x R 2R 3Q R x = 2 , the charge that has increased 3Q Q Q 2R = – (–Q)= hence charge flows into the Earth = NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 22 2 Example An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high resistance wire. After a long time what is the amount of heat loss ? Solution When two conducting spheres of equal radius are connected charge is equally distributed on them. So we can say that heat loss of system Q2 Q2 / 4 Q2 / 4 Q2 0 80R H = U – U 80R 8 0 R 1 6 0 R i f 42 E
JEE-Physics SOME WORKED OUT EXAMPLES Example#1 For a spherically symmetrical charge distribution, electric field at a distance r from the centre of sphere is E kr7ˆr , where k is a constant. What will be the volume charge density at a distance r from the centre of sphere ? (A) = 9k0r6 (B) = 5k0r3 (C) = 3k0r4 (D) =9k0r0 Solution Ans. (A) By using Gauss law q E 4r2 4r2dr ussian Surface 0 E dS 0 dr NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 O (Note : Check dimensionally that r6) Ga 4r2 dr 0 k0r9 kr7 4r2 r2 dr Example#2 Two positrons (e+) and two protons (p) are kept on four corners of a square of side a as shown in figure. The mass of proton is much larger than the mass of positron. Let q denotes the charge on the proton as well as the positron then the kinetic energies of one of the positrons and one of the protons respectively after a very long time will be– e+ p p e+ q2 1 q2 1 q2 q2 (A) 4 0 a 1 2 2 , 4 0 a 1 2 2 , (B) 2 0 a4 2 0 a q2 q2 q2 1 q2 , (D) 2 0 a 1 4 2 , 8 2 0 a (C) 4 0 a 4 0 a Solution Ans. (D) As mass of proton >>> mass of positron so initial acceleration of positron is much larger than proton. Therefore positron reach far away in very short time as compare to proton. e+ a p ap a aa a p a e+ pa 2K e 4kq2 2kq2 kq2 K e q2 1 1 2K kq2 0 Kp q2 a 2 a 2 0 2 and a2 2 0 a a 2 a 4 p 8 E 43
JEE-Physics Example#3 Four charges are placed at the circumference of a dial clock as shown in figure. If the clock has only hour hand, then the resultant force on a charge q0 placed at the centre, points in the direction which shows the time as :– +q 12 q 9 q0 3 +q 6 q (A) 1:30 (B) 7:30 (C) 4:30 (D) 10:30 Solution Ans. (B) +q q 9 q0 +q 7.30 Fnet 6 q Example#4 A small electric dipole is placed at origin with its dipole moment directed along positive x-axis. The direction of electric field at point (2, 22,0) is (A) along z-axis (B) along y-axis (C) along negative y-axis (D) along negative z-axis Solution Ans. (B) y (2,22) x q +q tan y 2;cot 1 Also tan tan 1 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Electrostatics\\English\\Theory.p65 x 2 2 2 cot + = 90° i.e., E is along positive y-axis. Example#5 Uniform electric field of magnitude 100 V/m in space is directed along the line y= 3 + x. Find the potential difference between point A (3, 1) & B (1,3). (A) 100 V (B) 2002V (C) 200 V (D) zero Solution Ans. (D) Slope of line AB = 3 1 1 which is perpendicular to direction of electric field. 13 44 E
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