physics part 1

Chaoptnertents PHYSICS www.notesdrive.com CBSE–XII PART – I Syllabus Unit-I 1-50 51-74 1. Electric Charges & Fields 75-104 2. Electrostatic Potential & Capacitance 105-148 149-162 Unit-II 3. Current Electricity Unit-III 4. Moving Charges and Magnetism 5. Magnetism and Matter Unit-IV 6. Electromagnetic Induction 7. Alternating Current Unit-V 8. Electromagnetic Waves

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ALLENÒ CBSE I SYLLABUS 2022 - 23 CLASS-XII SUBJECT-PHYSICS Units Topics No. of Periods Marks Unit - 1 Electrostatics 26 16 Chapter–1: Electric Charges and Fields 17 Chapter–2: Electrostatic Potential and Capacitance 18 12 Unit - 2 Chapter–3: Current Electricity 18 7 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\01_Syllabus.docxUnit - 3 Magnetic Effects of Current and Magnetism25 www.notesdrive.comChapter–4: Moving Charges and Magnetism Chapter–5: Magnetism and Matter Unit - 4 Electromagnetic Induction and Alternating Currents 24 Chapter–6: Electromagnetic Induction Chapter–7: Alternating Current Unit - 5 Chapter–8: Electromagnetic Waves 04 Unit – 6 Optics 30 Chapter–9: Ray Optics and Optical Instruments Chapter-10: Wave Optics Unit – 7 Chapter–11: Dual Nature of Radiation and Matter 8 Unit – 8 Atoms and Nuclei 15 Chapter–12: Atoms Chapter–13: Nuclei Unit – 9 Electronic Devices 10 Chapter–14: Semiconductor Electronics Total 160 70 E

II Physics ALLENÒ Unit I: Electrostatics Chapter–1: Electric Charges and Fields Electric charges, Conservation of charge, Coulomb's law-force between two-point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field. Electric flux, statement of Gauss's theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside) Chapter–2: Electrostatic Potential and Capacitance www.notesdrive.com Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\01_Syllabus.docx electric dipole in an electrostatic field. Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor (no derivation, formulae only). Unit II: Current Electricity Chapter–3: Current Electricity Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm's law, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, temperature dependence of resistance, Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff's rules, Wheatstone bridge. Unit III: Magnetic Effects of Current and Magnetism Chapter–4: Moving Charges and Magnetism Concept of magnetic field, Oersted's experiment. Biot - Savart law and its application to current carrying circular loop. Ampere's law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields. Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; Current loop as a magnetic dipole and its magnetic dipole moment, moving coil galvanometer-its current sensitivity and conversion to ammeter and voltmeter. E

ALLENÒ CBSE III Chapter–5: Magnetism and Matter Bar magnet, bar magnet as an equivalent solenoid (qualitative treatment only), magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis (qualitative treatment only), torque on a magnetic dipole (bar magnet) in a uniform magnetic field (qualitative treatment only), magnetic field lines. Magnetic properties of materials- Para-, dia- and ferro - magnetic substances with examples, Magnetization of materials, effect of temperature on magnetic properties. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\01_Syllabus.docxUnit IV: Electromagnetic Induction and Alternating Currents Chapter–6: Electromagnetic Induction www.notesdrive.com Electromagnetic induction; Faraday's laws, induced EMF and current; Lenz's Law, Self and mutual induction. Chapter–7: Alternating Current Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, Transformer. Unit V : Electromagnetic Waves Chapter–8: Electromagnetic Waves Basic idea of displacement current, Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses. Unit VI : Ray Optics Chapter–9: Ray Optics and Optical Instruments Ray Optics: Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism. Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers. Chapter–10: Wave Optics Wave Optics :- Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young's double slit experiment and expression for fringe width (No derivation final expression only), coherent sources and sustained interference of light, diffraction due to a single slit, width of central maxima (qualitative treatment only). E

IV Physics ALLENÒ Unit VII: Dual Nature of Radiation and Matter Chapter–11: Dual Nature of Radiation and Matter Dual nature of radiation, Photoelectric effect, Hertz and Lenard's observations; Einstein's photoelectric equation-particle nature of light. Experimental study of photoelectric effect Matter waves-wave nature of particles, de-Broglie relation. Unit VIII: Atoms and Nuclei Chapter–12: Atoms Alpha-particle scattering experiment; Rutherford's model of atom; Bohr model of hydrogen atom, Expression for radius of nth possible orbit, velocity and energy of electron in its orbit, hydrogen line spectra (qualitative treatment only). www.notesdrive.com Chapter–13: Nuclei Composition and size of nucleus, nuclear force node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\01_Syllabus.docx Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion. Unit IX: Electronic Devices Chapter–14: Semiconductor Electronics : Materials, Devices and Simple Circuits Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors- p and n type, p-n junction Semiconductor diode - I-V characteristics in forward and reverse bias, application of junction diode - diode as a rectifier. E

ALLENÒ CBSE 1 UNIT - I : ELECTROSTATICS CHAPTER-1 : ELECTRIC CHARGES & FIELDS 1. ELECTRIC CHARGE Charge is a property of substance due to which it shows electric and magnetic effects. Kinds of charges : positive and negative. S.I. unit of charge is coulomb (C). Basic property of charge : (a) Additivity (b) Conservation (c) Quantisation Mode of charging : (a) charging by friction, (b) charging by conduction, (c) charging by induction Note : Electric charge is invariant because value of electric charge does not depend on frame of reference. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 2. COULOMB'S LAW www.notesdrive.com The mutual electrostatics force between two point charges (q1 and q2) is proportional to the product (q1q2) and inversely proportional to the square of the distance (r2) separating them. The force acts along the line joining two charges. Force in free space (air / vacuum) Fr = 1 q1q2 ˆr q1 q2 4p Î0 r2 r Force in any medium Fr = 1 q1q2 ˆr 4p Î r2 where e = e0er Îr = relative permittivity, Î = permittivity of medium, Î0 = permittivity of free space (air / vacuum) = 8.8 × 10–12 C2N–1m–2 NOTE :The Law is applicable only for static and point charges. Moving charges may result in magnetic interaction and if charges are spread on bodies then induction may change the charge distribution. r E 3. ELECTRIC FIELD / ELECTRIC INTENSITY / ELECTRIC FIELD STRENGTH ( ) It is defined as the net force on unit positive charge due to all other charges. It is vector quantity. Er = Fr unit is N/C or V/m. q (a) Electric field due to point charge Er = kq ˆr r2 P E r r q E

2 Physics ALLENÒ 4. ELECTRIC FLUX : f = ò Er.dAr SI unit = N·m2 C (i) For uniform electric field; f = rr = E A co s q where q = angle between Er & area vector ( Ar ) . E .A Flux is contributed only due to the component of electric field which is perpendicular to the plane. (ii) If Er is not uniform throughout the area A , then f = òEr.dAr (iii) r represent area vector normal to the surface and pointing outwards from a closed surface. dA 5. GAUSS’S LAW www.notesdrive.com It states that flux of electric field through any closed surface S is 1 times the total charge node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx e0 enclosed by S. Ñò Er.drs = åq (Applicable only to closed surface) Î0 Ñòf = Er × d Ar = qen e0 where qen= net charge enclosed by the closed surface. The above closed surface is called gaussian surface. f does not depend on the : (i) Shape and size of the gaussian surface (ii) The charges located outside the gaussian surface. (iii) Location of charge inside the gaussian surface. Application of Gauss's Law (i) Field due to an infinitely long straight uniformly charaged wire : Er = l nˆ 2pe0 ++ Cylindrical Gaussian ++ surface + |E| + + O + + ++ P r + + dA E ++ ++ ++ + r + + ++ + E

ALLENÒ CBSE 3 (ii) Field due to uniformly charged infinite plane non conducting sheet. Er = s nˆ 2e0 E + + E + + s ++ + + ++ r n^ P + + + r + + + s (iii) Electric field due to thin spherical shell. (a) outside shell ® E = kq R r2 Or (b) on the shell ® E =node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxkqP R2 www.notesdrive.com (b) inside the shell ® E = 0 So, charge on spherical shell behave as charge concentrated at centre of sphere for points situated outside. 6. ELECTROSTATIC FIELD LINES An electrostatic field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electric field at that point. AB qA>qB Field lines of electrostatic field have following properties : (i) Imaginary lines (ii) Never intersect each other (iii) Electrostatic field lines never forms closed loops (iv) Field lines ends or starts normally at the surface of a conductor. (v) If there is no electric field there will be no field lines. (vi) Number of electric field lines per unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field while distant lines weak field. E

4 Physics ALLENÒ 7. ELECTRIC DIPOLE If equal and opposite point charges are placed at very small separation then system is known as dipole. 2a q –q Electric dipole moment ( pr ) It is defined as the product of magnitude of either charge and distance between them. pr = q ·2ar Electric dipole moment : (i) It is a vector quantity, (ii) Direction is from –ve to +ve charge. Electric dipole in uniform electric fieldwww.notesdrive.com 1. Torque tr = pr × Er or t = pE sin q node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 2. Work done in rotating dipole from q1 to q2 angle in external electric field. W = pE(cosq1–cosq2) 3. Potential energy = –pE cosq Note : • In uniform field, force on a dipole = 0, torque may or may not be zero. • In general in non-uniform field, force on a dipole ¹ 0 and torque may or may not be zero • In non uniform electric field Þ Fe = pr. uduEr dr • When Pr is parallel to Er then the dipole is in stable equilibrium • When Pr is antiparallel to Er then the dipole is in unstable equilibrium Electric field and potential due to dipole Electric field Potential 1. at axial 2kp kp r3 r2 2. at equatorial kp 0 r3 3. at general position kp 1 + 3cos2 q kpcos q r3 r2 E

ALLENÒ CBSE 5 CHAPTER-2 : ELECTROSTATIC POTENTIAL & CAPACITANCE 1. ELECTROSTATIC POTENTIAL It is the work done against the field to take a unit positive charge from infinity (reference point) to the given point without gaining any kinetic energy. VP = é ( W -P )ext ù ëê ¥ ûú q SI unit = volt (V) Potential at a point due to positive charge is positive & due to negative charge is negative. Potential Due to Special Charge Distribution : (i) Point charge V = kq r node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxVr qp www.notesdrive.com r (ii) Charged conducting sphere (a) VC kq ; r>R V BC =r Kq A R (b) VB = kq ; r=R R R R r kq (c) VA = R ; r<R 2. POTENTIAL DIFFERENCE The potential difference between two points A and B is work done by external agent against electric field in taking a unit positive charge from B to A without acceleration (or keeping Kinetic Energy constant) . VA - VB = (WBA )ext q Potential difference between two points in an electric field does not depend on the path between them. 3. ELECTRIC POTENTIAL ENERGY OF TWO CHARGES It is the amount of work required to bring the two point charges to a particular separation from infinity without change in kinetic energy. U = 1 q1q2 4p Î0 r 4. EQUIPOTENTIAL SURFACE AND EQUIPOTENTIAL REGION In an electric field the locus of points of equal potential is called an equipotential surface. An equipotential surface and the electric field line meet at right angles. The region where E = 0, Potential of the whole region must remain constant as no work is done in displacement of charge in it, it is called as equipotential region like conducting bodies. E

6 Physics ALLENÒ 5. POTENTIAL GRADIENT | Er | = - dV dr Electric potential goes on decreasing when we go in the direction of electrostatic field. 6. ELECTROSTATICS OF CONDUCTORS • Electric field in the bulk of the conductor (volume) is zero while it is perpendicular to the surface in electrostatics. • Excess charge resides on the free surface of conductor in electrostatic condition. • Potential throughout the volume of the conductor is same in electrostatics. • Electric field at the surface of a charge conductor is given as Er = s nˆ e0 Electrostatics Shielding • A metallic shield in form of a hollow conducting shell built to block an electric field because in a hollow conducting shell, the electric field is zero at every point. This method of protection is called electrostatics shielding. www.notesdrive.com 7. DIELECTRICS AND POLARISATION node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx Dielectrics are non conducting substances. A dielectric (whether polar or non polar) develops a net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation (P). rr P = ceE Dielectric strength of material The minimum electric field required to ionize the medium or the maximum electric field which the medium can bear without break down. Dielectric strength of air = 3 × 106 V/m. 8. CAPACITOR & CAPACITANCE A capacitor consists of two conductors carrying charges of equal magnitude and opposite sign. The capacitance C of any capacitor is the ratio of magnitude of the charge Q on either conductor to the potential difference V between them Þ C = Q V unit of capacitance = farad (F). If an isolated conductor is brought near to charged conductor then it increases the capacity of system. This is the principle of capacitor. The capacitance depends on the geometry of the conductors & presence of dielectric medium. It doesn't depend on the charge of conductor or potential difference. Capacitance of an isolated Spherical Conductor C = 4p Î0Îr R in a medium C = 4p Î0 R in air E

ALLENÒ CBSE 7 8.1 Parallel Plate Capacitor (i) Uniform Di-electric Medium : If two parallel plates each of area A & separated by a distance d are charged with equal & opposite charge Q, then the system is called a parallel plate capacitor & its capacitance is given by, C = Î0 Îr A in a medium; C = Î0 A with air as medium d d This result is only valid when the electric field between plates of capacitor is constant. (ii) Medium Partly Air : C = Î0 A t ö = Î0 A t æ Îr ÷ (d - t) + d - ç t - ø Îr d t è When a di-electric slab of thickness t & relative permittivity Îr is introduced between the plates of an air capacitor, then Îr Î0 the distance between the plates is effectively reduced by node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx æ t t ö irrespective of the position of the di-electric slab. On www.notesdrive.comç-÷ è ø Îr filling the space between the plates of a parallel plate air capacitor with a dielectric, capacity of the capacitor is increased because the same amount of charge can be stored at a reduced potential. (iii) Composite Medium : (a) When dielectrics are in series d Îr1 Îr2 Îr3 C= Î0 A t1 + t2 + t3 Î Î Îr1 r2 r3 (b) When dielectrics are in parallel d A1, Îr1 t1 t2 t3 A2, Îr2 A3, Îr3 C= e0 [A1Îr1 + A2Îr2 + A3Îr3] d 8.2 Combination Of Capacitors (i) Capacitors in Series : In this arrangement all the capacitors when QQ Q uncharged get the same charge Q but the potential difference across each will differ if the C1 C2 C3 capacitance are unequal. In series combination the potential difference on capacitor will be inversely V1 V2 V3 proportional to its capacitance. 1 = 1 + 1 + 1 + ..... + 1 Ceq. C1 C2 C3 Cn Q = (Ceq.V) = C1V1 = C2V2 = C3V3 E

8 Physics ALLENÒ (ii) Capacitors In Parallel : When one plate of each capacitor is connected to the Q1 C1,V positive terminal of the battery & the other plate of each Q2 C2,V capacitor is connected to the negative terminal of the battery, then the capacitors are said to be in parallel connection. Q3 C3,V The capacitors have the same potential difference, V but QV the charge on each one is different (if the capacitors are unequal). In parallel combination charge on capacitor will be distributed in the ratio of their capacitance. Ceq. = C1 + C2 + C3 + ...... + Cn . Q = Q1 + Q2 + Q3 V= Q = Q1 = Q2 = Q3 C eq C1 C2 C3 www.notesdrive.com 8.3 Energy Stored in a charged capacitor node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx Capacitance C, charge Q & potential difference V; then energy stored is U = 1 CV2 = 1 QV = 1 2 22 Q2 where Q &V are charge & voltage of the capacitor C. C This energy is stored in the electrostatic field set up in the di-electric medium between the conducting plates of the capacitor. Energy density u = 1 E2, E is electric field between plate of capacitor. 2 Î0 Work done by battery to charge a capacitor W = CV2 = QV = Q2 C 8.4 Sharing of Charges When two charged conductors of capacitance C1 & C2 at potential V1 & V2 respectively are connected by a conducting wire, the charge flows from higher potential conductor to lower potential conductor, until the potential of the two condensers becomes equal. The common potential (V) after sharing of charges; (i) When the positive plate of one connected to positive plate of other then common potential. V= net ch arg e = q1 + q2 = C1V1 + C2V2 . net capaci tan ce C1 + C2 C1 + C2 Charges after sharing q1 = C1V & q2 = C2V. In this process energy is lost in the connecting wire as heat. U - Uinitial = C1 C2 (V1 - V2)2 C1 + C2 ( )This loss of energy is final 2 (ii) When positive plate of one connected with negative of other, then common potential. VC = C1V1 - C2V2 . C1 + C2 After connection ratio of charge would be in ratio of capacitance q1 = C1 q2 C2 E

ALLENÒ CBSE 9 NCERT IMPORTANT QUESTIONS CHAPTER - 1 : ELECTRIC CHARGES AND FIELDS 1. How much positive and negative charge is there in a cup of water (250 gram)? Sol. The molecular mass of water is 18g. Thus, one mole (= 6.02 × 1023 molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250/18) × 6.02 × 1023. Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to (250/18) × 6.02 × 1023 × 10 × 1.6 × 10–19 C = 1.34 × 107 C. 2. Four point charges qA = 2mC, qB = –5mC, qC = 2mC, and qD = –5mC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1mC placed at the centre of the node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx square? www.notesdrive.com Sol. FO A = k (2µC)(1µC) D C AO 2 –5µC +2µC FO B = k(5µC) (1µC) O BO2 FO C = k(2µC) (1µC) CO2 A+2µC –5µC B FO D k(5µC) (1µC) = DO2 In a square AO = BO = CO = DO So FOA = FOC FOB = FOD These force are same in magnitude but opposite in direction so resultant force will be equal to zero. FR = 0. 3. A system has two charges qA = 2.5 × 10–7C and qB = – 2.5 × 10–7C located at points A : (0, 0, –15 cm) and B : (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system? Sol. Total charge = 2.5 × 10–7 – 2.5 × 10–7 = 0 Electric dipole moment is pr = q(2ar) = 2.5 × 10–7 × (0.15 + 0.15) C–m = 7.5 × 10–8 C–m The direction of dipole moment is along –Z Axis. E

10 Physics ALLENÒ 4. The electric field components in figure are Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. Sol. (a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and AS is ± p/2. Therefore, the flux f = E.DS is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is EL = ax1/2 = aa1/2(x = a at the left face). The magnitude of electric field at the right face is ER = ax1/2 = a (2a)1/2 (x = 2a at the right face). The corresponding fluxes are fL = EL. DS = DS EL . nˆ L = EL DS cos q = – EL DS, since q = 180° = –ELa2 fR = ER.DS = ER DS cosq = ER DS, since q = 0° = ERa2 Net flux through the cube = fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2] = aa5/2 ( 2 - 1) = 800 (0.1)5/2 ( 2 - 1) = 1.05 N m2 C–1 (b) We can use Gauss’s law to find the total charge q inside the cube. We have f = q/e0 or q = feo. Therefore, q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C. 5. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E : (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? +s –s www.notesdrive.com Sol. I node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx III II PQ E

ALLENÒ CBSE 11 (a) in the outer region of the first plate E1 = sP + sQ = 17 ´10-22 – 17 ´10-22 ; E1 = 0 2e0 2e0 (i.e. Electric field is zero.) (b) Similarly, electric field is zero in this case also (c) between the plates EIII = sP - sQ 2e0 = 17 ´10-22 + 17 ´10-22 2e0 34 ´10-22 2 ´8.854 ´10-12 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxEIII == 1.92 × 10–10 NC–1 www.notesdrive.com6. Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole of dipole moment and length 2a. What is the direction of this field ? Sol. EB EBsinq EBcosq q P x¢ q EAcosq EA EAsinq r –q q O q +q A B 2a We consider a dipole consisting of –q and +q separated by a distance 2a. Let P be a point on the equatorial line. Er A = 1 q along uuur 4pe0 PA (AP)2 1 q 4pe0 r2 + a2 ( )EA = Er B = 1 q along uuur 4pe0 (BP)2 BP 1 q 4pe0 r2 + a2 ( )EB = E

12 Physics ALLENÒ The resultant intensity is the vector sum of the intensities along uuur and uuur . EA and EB can be PA BP resolved into vertical and horizontal components. The vertical components of EA and EB cancel each other as they are equal and oppositely directed. So the horizontal components add up to give the resultant field. E = EA cos q + EB cos q E = 2EAcosq , as EA = EB Substituting, cosq = a 1 in the above equation ( )r2 + a2 2 = 2EAcosq 2 q a 4pe0 r2 + a2 1 r2 + a2 2 ( ) ( )E = E=www.notesdrive.comkp along uuur (As p = q × 2a) BA node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx3 ( )r2 + a2 2 As a special case, If a2 << r2 then, E= kp along uuur r3 BA Electric field intensity at an axial point is twice the electric intensity on the equatorial line. Direction of field will be against the direction of dipole moment. 7. Derive an expression for the electric field E due to a dipole of length '2a' at a point distant r from the centre of the dipole on the axial line. Sol. Let consider a dipole system, Aa aB EA ®P –q Op +q XX EB ®EA r ®EB Here, AO = OB = a OP = r BP = r – a AP = r + a r Elec. field ( EB ) , due to charge at point 'B' is towards 'P' Elec. field (Er A ) , due to charge at point 'A' is opposite to 'P' Now, according to the superposition principle, rr rr E P = E axial = E A + E B Ep = -kq + kq (r + a)2 (r - a)2 = é 1 - (r 1 ù kq ê ú - a)2 + a)2 ûú ëê ( r E

ALLENÒ CBSE 13 = kq é ( r + a + r -a) ( r + a - r + a ) ù ê - a)2 ( r + a ú êë (r )2 úû kq( 2r )(2a ) ( )= r 2 - a 2 2 2rk(2a)(q) r2 - a2 2 ( )r = rˆ Q| pr |= 2a ´q Eaxial = 2kp rr if a2 << r2 r4 2kpr then r = r3 E axial It will be directed in the direction of electric dipole moment, pr node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 8. Use Gauss's theorem to find the electric field due to a uniformly charged infinitely large plane www.notesdrive.com thin sheet with surface charge density s. A nn Sol. dEsinq dE E,n E,n q 2dEcosq ds q dEsinq dE r As shown in figure, considering a cylindrical gaussian surface of cross section A (i) Flux through curved surface : f = ò Er.dSr =ò Eds cos 90º = 0 At the points on the curved surface, the field vector E and area vector dS make an angle 90o with each other. Therefore, curved surface does not contribute to the flux. Flux through end caps : ur r f = Ñò E.dS = Ñò EdScos 0° = EA Hence, the total flux through the closed surface is : f = Flux through both end caps + flux through curved surface or f = EA + EA + 0 = 2EA (1) Now according to Gauss’ law for electrostatics f = q/e0 (2) Comparing equations (1) and (2), we get 2EA = q/e0 E = q/2e0A (3) E

14 Physics ALLENÒ The area of sheet enclosed in the Gaussian cylinder is also A. Therefore, the charge contained in the cylinder, q = sA as s (surface charge density) = q/A Substituting this value of q in equation (3), we get E = sA/2e0A or E = s/2e0 This is the relation for electric field due to an infinite plane sheet of charge. The field is uniform and does not depend on the distance from the plane sheet of charge. 9. Use Gauss' law to derive the expression for the electric field due to a straight uniformly charged infinite line of charge density l C/m. Sol. To calculate the electric field, imagine a cylindrical Gaussian www.notesdrive.comsurface, since the field is everywhere radial, flux through two ends node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxof the cylindrical Gaussian surface is zero.ur E At cylindrical part of the surface electric field is normal to the surface at every point and its magnitude is constant. Therefore flux through the Gaussian surface. = Flux through the curved cylindrical part of the surface. = E × 2prl ......(i) Applying Gauss's Law Flux f = q enclosed e0 Total charge enclosed = Linear charge density × l = ll \\ f = ll .......(ii) e0 Using Equations (i) & (ii) E × 2 p rl = ll Þ Er = l nˆ e0 2pe0r (where nˆ is a unit vector normal to the line charge) E

ALLENÒ CBSE 15 NCERT IMPORTANT QUESTIONS CHAPTER-2 : ELECTROSTATIC POTENTIAL AND CAPACITANCE 1. Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Ax (15–x)cm B q1 =3×10–8C = –2×10–8C Sol. P q2 cm 15 Let at point P potential is zero VP = V1 + V2 = 0 Vp = kq1 + kq2 = 0 X (15 - X) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx Vp =k é3´10-8 - 2´10-8 ù = 0 3 ´10-8 2 ´10-8 www.notesdrive.comê ú Þ X = ë X (15 - X) û (15 - X ) Þ 45 – 3X = 2X Þ X = 9cm 2. A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? Sol. C = Îo A t - t) + K (d = Îo A 3d 3 øö÷ + 4K çèæ d - 4 d = Îo A Here æ C o = Îo A ö èç d ø÷ d çèæ1 + 3 ö 4 K ø÷ C = 4KCo K+3 3. A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Sol. Energy loss DU = 1 C1C2 (V1 - V2 )2 2 Cl + C2 = 1 600 ´ 600 ´10-24 (200 – 0)2 2 (600 + 600) ´10-12 = 1 ´ 600 ´ 600 ´ 10 -12 ´ 4 ´ 104 2 1200 = 6 × 10–6 J E

16 Physics ALLENÒ 4. Derive an expression for the work done in rotating a dipole from the angle q0 to q1 in a uniform electric field . Sol. +q A +qE –qE 2a q B –q As we know that, when a dipole is placed in a uniform electric field, net force on dipole is zero but it experiences a torque, which can be given as, tr = pr ´Er This torque rotates the dipole unless it is placed parallel or anti-parallel to the external field. If we apply an external and opposite torque, it neutralizes the effect of this torque given by text and it rotates the dipole from the angle q0 to an angle q1 at an infinitesimal angular speed without any angular acceleration. www.notesdrive.com The amount of work done by the external torque can be given by node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx q1 q1 ò òW = text dq = pE sin q dq q0 q0 W = pE [- cos q]q1 = pE (cos q0 - cos q1 ) q0 5. (a) Determine the electrostatic potential energy of a system consisting of two charges 7mC and –2mC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? Sol. (a) U= 1 q1 q2 4pe0 r U = 9´109 ´ 7 ´ (-2) ´10-12 = 0.7J 0.18 (b) W = U2 – U1 = 0 – U = 0 – (–0.7) = 0.7 J 6. Write the definition of electric potential. Calculate the electric potential due to a point charge Q at a distance r from it. Draw a graph between electric potential V and distance r for a point charge Q. Sol. Electric potential :- The amount of work needed to bring a unit positive charge from infinity to a specific point inside the electric field of a positive charge is called electric potential at that point. Potential due to point charge :- electrostatic force on + q0 at point A due to charge Q Q +q0 B FFeexxtt.A FFee ¥ Or P A dx B +q0 x Fe = 1 Qq0 4 pe 0 x2 E

ALLENÒ CBSE 17 Work done in moving a charge +q0 in short displacement from A to B dW = Fedx =– Qq 0 ´ 1 dx 4pe0 x2 Total work done in moving a charge q0 from ¥ to r - òW = – Qq0 r 1 dx 4pe0 ¥ x2 W = - Qq0 ëêé- 1 ù r 4pe0 x ûú ¥ W= Qq0 é1 - 1 ùr ÞW= Qq 0 4pe0 ëê r ¥ úû ¥ 4pe0 r from definition of potential node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx V= W ÞV= 1 Q V www.notesdrive.com q0 4pe0 r V = kQ Þ V µ 1 V µ 1r rectangular r r Hyperbola graph between V and r r 7. What are equipotential surface ? Write any two properties of equipotential surfaces .Show equipotential surface due to – [1] a point charge , [2] around a dipole Sol. Equipotential surface is a surface which has equal potential on every point of it. (a) The net work done in moving a charge from one point to another on this surface is zero. (b) Electric field is in the direction in which the potential decreases steepest. [1] A point charge v3 Equipotential surface v2 d2 > d1 v1 q dd1 1 dd2 2 [2] Around a dipole E

18 Physics ALLENÒ 8. Define electric capacity. Derive an expression for the capacitance of a parallel plate capacitor in which a dielectric medium of dielectric constant K fills the space between the plates. Draw the necessary diagram. Sol. Electrical capacitance : The ability of a conductor to store electrical energy (or charge) is called electric capacitance. Charge which is given to a conductor is directly proportional to it's increasing potenital. q µ V Þ q = CV Þ C = q V C Þ electrical capacity of a conductor. Capacitance of parallel plates capacitor :- electric field intensity between two plates of capacitor. Em = s www.notesdrive.comìs=q e ï A node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxí îïe = e0K Em = q e0KA Potential difference between two plates of capacitor V = Em × d V = qd e0KA +q d –––––––––––––––q + + + + + + + + + Plate Area Þ A dielectric constant Þ K Potential difference C = q ÞC= æ q öÞ C = e0KA V qd d çè e0KA ÷ø 9. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? Sol. C = 12 pF = 12 × 10–12 F V = 50 volt U=? U = 1 CV2 = 1 × 12 × 10–12 × 50 × 50 J 22 E

ALLENÒ CBSE 19 10. What does mean by energy stored in capacitor? Prove that \"The ratio of change in potentials of conductors after connecting two charged conductors is inversely proportional to their capacitances\". Sol. Energy stored in conductor :- Work has to be done in charging a capacitor against the force of repulsion by the already existing charges on it. This work is stored as a potential energy in the form of electric field of a capacitor. Let A and B are two isolated charged conductor. Their radius are R1 and R2 respectively and capacitance C1 and C2 respectively. When charge Q1 given to A and Q2 to B, then their potential becomes V1 and V2 respectively. (Q1, C+1, V+ 1) (Q2, +C2,+V2) (Q'1, C1, V) (Q'2, C2, V) + + +++ +++ A B R2 A B R2 R1 +++ ++ +++ +++ node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxR1 www.notesdrive.co++++ +++m+++++++ +++ +++ +++ Before connection After connection Before connection with conducting wire - Q1 = C1 V1 and Q2 = C2 V2 Total charge of both the conductor - Q = Q1 + Q2 Q = C1V1 + C2V2 ....(1) When both are connected by conducting wire of negligible capacitance then charge flows from high potential conductor to low potential conductor such that the final potential of two becomes same. Charges after redistribution- Q1' = C1 V and Q2' = C2 V Where V is common potential. From law of conservation of charges Q1 + Q2 = Q1' + Q2' C1V1 + C2V2 = C1V + C2V V = C1V1 + C2V2 ....(2) C1 + C2 difference in potential of conductor A DV1 = V1 – V DV1 = V1 – æ C1V1 + C2V2 ö ç C1 + C2 ÷ è ø DV1 = C 2 (V1 - V2 ) ....(3) C1 + C2 E

20 Physics ALLENÒ difference in potential of conductor B DV2 = V – V2 DV2 = æ C1V1 + C2V2 ö - V2 ç C1 + C2 ÷ è ø DV2 = C1 ( V1 - V2 ) ....(4) C1 + C 2 Equation (3) ¸ (4) DV1 = C2 Proved DV2 C1 CHAPTER - 1 : ELECTRIC CHARGES AND FIELDS CHAPTER - 2 : ELECTROSTATIC POTENTIAL AND CAPACITANCE EXERCISE-I ONE MARK QUESTIONS : 1. Draw the pattern of electric field lines due to an electric dipole. www.notesdrive.com Sol. + – node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 2. Draw equipotential surfaces for an electric dipole. Sol. 3. Define dielectric constant of a medium. What is it’s S.I. unit ? Sol. Dielectric constant (or relative permittivity) of a medium is the ratio of the absolute permittivity of a medium to the permittivity of free space. K ( or er ) = e/e0 Value of K is more than 1 for any dielectric medium. As it is a ratio so it has no unit. E

ALLENÒ CBSE 21 4. Two equal balls having equal positive charge ‘q’ coulomb are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two ? Sol. Effective force will be reduced.* Force between point charges is given by following formula Fm = Fo (for plastic Îr >1) Îr So, Fm < Fo Comment :- In board it is assumed that sheet is sufficient thick to fill the space between balls. 5. Why the electrostatic field line do not form closed loop ? Sol. The field lines start (or diverge) from the positive charge and terminate (or converge) on negativenode06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx www.notesdrive.comcharge. If electric field lines form closed loops, then these lines must originate and terminate on the same charge which is not possible So, electric field lines can't form closed loop . OR Electrostatic field is conservative in nature. 6. A point charge +Q is placed in the vicinity of a conducting surface. Trace the field lines between the charge and the conducting surface. Sol. –+ –+ +Q – + –+ –+ 7. The figure shows the field lines of a positive point charge. What will be the sign of the potential energy difference of a small negative charge between the points Q and P ? Justify your answer. •Q + •P Sol. P.E. of any charge configuration is given by – U = kq1q2/r charge q2 is negative so P.E of configuration = –kq1q2/r Point Q is at larger distance thus potential energy is greater as compared to point P, so potential energy difference is positive. [Refer solved examples 2.3, page no. 59 NCERT] E

22 Physics ALLENÒ 8. Depict the electric field lines due to two positive charges kept at certain distance apart. Sol. 9. Why electric field lines are perpendicular at a point on an equipotential surface of a conductor ? Sol. At equipotential surface Work done, W=0 So, Fr.drr = 0 –qo Eur.drr = 0 Er ^ d rr www.notesdrive.com E ^ E.P.S. (Equipotential Surface) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx Hint : Let we consider a hypothetical condition in which the perpendicular relation does not hold. If E were not perpendicular to equipotential surface, then there would be a component of E parallel to the surface. This field component would exert an electric force on a test charge placed on that surface. As the charge moves along the surface ,work would be done by this component of the electric force. Which causes potential to change. This would violates the nature of equipotential surface, on which work done is zero so angle is 90o between electric force and displacement. Thus electric field lines must be (which is direction of electric force) perpendicular at a point on an equipotential surface. 10. Define electric flux Write its SI Unit. Sol. The number of electric field lines passing through given area is called electric flux. f = A (Ecosq) A® q E® f = Er ×Ar Electric Field Electric flux through an area is the dot product of magnitude of Er and Ar . The S.I. Unit of ‘electric flux’ is N-m2C–1 or V-m. 11. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when it’s radius is increased ? Sol. We know that flux through the Guassian surface is given by f = q/e0 As flux is independent of radius, so it is not affected on changing the radius. [r > R] E

ALLENÒ CBSE 23 12. A point charge +Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero ? +Q O• A• B• Sol. We know that V = kQ/r \\ Vµ 1 r So VA > VB VA – VB will be positive. 13. What is the amount of work done in moving a point charge Q around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located ? Sol. Work done to move the charge over the circular arc is zero, because Q is moving over an equipotential circular arc. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 14. The field lines of a negative point charge are shown in the figure. Does the kinetic energy of a small negative charge increase or decrease in going from B to A ?www.notesdrive.com –q •A •B Sol. When we move –ve charge from B to A, repulsion will increase So Kinetic energy decrease. 15. A charge ‘q’ is moved from point A above a dipole to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process. A –q• +•q B Sol. The potential due to a dipole at any point on equatorial plane is zero. That's why work done is zero in the above process . As work done (W)= charge × change in potential (W = q × D v ) = 0 16. Write a relation for polarisation ( Pr ) of a dielectric material in the presence of an external electric field Er . ur ur Sol. P = ce E , ce represent electrical susceptibility of dielectric material represents degree of polarization of a dielectric material in response to an applied external electric field. E

24 Physics ALLENÒ 17. The given graph shows variation of charge ‘q’ versus potential difference ‘V’ for two capacitors C1 and C2. Both the capacitors have same plate separation but plate area of C2 is greater than that of C1. Which line (A or B) corresponds to C1 and why ? A qB V Sol. We know that Slope of q-v graph is capacitance, so capacitance for graph-A > Capacitance for graph-B. According to question A2 > A1 (Area) C2 > C1 So graph-A corresponds to C2. www.notesdrive.com EXERCISE-II node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx TWO MARK QUESTIONS : 1. Derive an expression for the work done in rotating a dipole from the angle q0 to q1 in a uniform electric field. (Refer to Ch-2, NCERT Q.4) 2. What is electrostatic shielding ? How is this property used in actual practice ? Is the potential in the cavity of a charged conductor zero ? Sol. Electrostatic shielding is the method of protecting a certain region of space from external electric field, as the electric field inside a conductor is zero. Therefore to protect electrostatically sensitive instruments from external electric field, we enclose them in a hollow conductor. Inside hollow conductor instrument does not get affected (damage) from external field. For example we use a co-axial cable to transfer video signal without external interference Electric potential in the cavity of a charged conductor is not zero. It is constant throughout the volume of the conductor and has the same value as on its surface. EXERCISE-III THREE MARK QUESTIONS : 1. In a network, four capacitors C1, C2, C3 and C4 are connected as shown in the figure. C2 = 6µF C4 C3 12 µF 4 µF C1 = 3µF 8V (a) Calculate the net capacitance in the circuit. (b) If the charge on the capacitor C1 is 6 µC, (i) calculate the charge on the capacitors C3 and C4, and (ii) net energy stored in the capacitors C3 and C4 connected in series. E

ALLENÒ CBSE 25 C4 C3 C2 = 6µF 12 µF 4 µF C1 = 3µF Sol. 8V (a) Equivalent capacitance of C1 & C2– C' = C1 + C2 C ' = 9mF Equivalent capacitance of circuit- 1 111 Ceq = C4 + C3 + C' 1 1 11 Ceq = 12 + 4 + 9 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 1 4 µF Þ Ceq = 9 mF www.notesdrive.comCeq=9 4 (b) Total charge given by battery- q = CeqV q= 9 ×8Þ q = 18mC 4 (i) Capacitors C3 and C4 are in series, so charge on both capacitors will be same & equal to the total charge supplied by battery. charge on C3 Þ q3 = 18µC charge on C4 Þ q4 = 18µC (ii) Net energy stored in the capacitors C3 and C4. U= q 2 + q 2 {q4 = q3} 3 4 2C3 2C4 U= q32 é1 + 1ù 2 ê ú ë C3 C4 û = 18 ´18 é C4 + C3 ù 2 ê C3 ´ C4 ú ë û U= 18´18 é 16 ù 2 êë12´ 4úû U = 54 µJ Þ U = 54 ´10-6 J E

26 Physics ALLENÒ 2. A 200 µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change? Sol. K= 10 d=5mm 5mm 10mm 100Vwww.notesdrive.com 100V (i) C0 = node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxÎ0 A=Î0 A...(i) d 5 When dielectric slab is placed. C= Îo A = Îo A 5 (d - t) + t - 5) + 10 (10 Îr C = 2 Îo A ...(ii) 11 Eq. (ii) divided by (i) C= 10 C 0 ...(ii) 11 (ii) E0 = s ...(i) Îo V = Ed E= V = Vo + Vm d d E = E o (10 - 5) + Em ´5 = 1 (Eo + Em ) 10 2 E = 1æ s + s ö = s çæè1 + 1 ö 2 èç Îo Îo K ø÷ 2 Îo K ø÷ E= Eo æçè1+ 1 ö = 11 Eo 2 10 ø÷ 20 (iii) Energy U = 1 CNetV2= 1 æ10 C0 ö (100)2 2 2 èç 11 ÷ø U= 1 ´ 10 × 200 × 10–6 × 104 2 11 Energy = 10 Joule 11 E

ALLENÒ CBSE 27 3. Two large charged plane sheets of charge densities s and –2s C/m2 are arranged vertically with a separation of d between them. Deduce expressions for the electric field at point (i) to the left of the first sheet, (ii) to the right of the second sheet, and (iii) between the two sheets. A B Sol. E1 ++++++s+++++ + E1 ––––––2s––––– – E2 E2 + E2 – E1 I + – + III – II + – Electric field in the region left to the first sheet node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxE = E1 – E2E1 =s;E2 =2s 2 Îo 2 Îo www.notesdrive.com EI = s - 2s 2e0 2e0 EI = – s 2e0 It is towards right Electric field in the region to the right of second sheet E = E1 – E2 EII = s - 2s 2e0 2e0 EII = - s 2e0 It is towards left Electric field between the two sheets EIII = E1 + E2 EIII = s + s e0 2e0 EIII = 3s 2e0 Electric field is towards right. 4. Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole of dipole moment pr and length 2a. What is the direction of this field ? (Refer to NCERT Q.6) E

28 Physics ALLENÒ 5. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity Er at a point on the axis of the ring. Hence show that for points at large distance from the ring, it behaves like a point charge. Sol. + dl + + + r dEsinq dE + + a+ q q 2dEcosq +x qq +O + + + + dEsinq dE + + dl Suppose that the ring is placed with its plane perpendicular to the x-axis, as shown in the abovewww.notesdrive.com diagram .We consider a small element dl of the ring. So the charge dq on the element dl is node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx dq = q dl æçèQ l = q ö l = charge per unit length 2 pa 2pa ø÷ \\ The magnitude of the field dEr produced by the element dl at the field point P is As shown in the figure, the field dE has two components : (i) the axial component Þ dEcosq (ii) the perpendicular component Þ dEsinq Since the perpendicular components of any two diametrically opposite elements are equal and opposite, they all cancel out in pairs. Only the axial components will add up to produce the resultant field at point P, which is given by, 2 pa E = ò dE cos q 0 2pa kq dl x kqx 2pa x 2pa r2 r 2par3 0 r òE = ó æ ö = dl éëêas cos q = ù õô çè ÷ø ûú 0 kqx 2pa kqx 1 2par3 0 2pa x2 + a2 3/2 ( )[ ] ( )E= l = 2pa [as r2 = x2 + a2] E= kqx 1 qx x2 + a2 3/2 = 4pe0 x2 + a2 3/2 ( ) ( )or Special case : For points at large distances from the ring x >> a \\ E = kq = 1 q x2 4pe0 x2 This is same as the field due to a point charge, indicating that for far off axial points, the charged ring behaves as a point charge. E

ALLENÒ CBSE 29 6. (a) Define torque acting on a dipole of dipole moment pr placed in a uniform electric field r . E Express it in the vector form and point out the direction along which it acts. (b) What happens if the field is non-uniform ? (c) What would happen if the external field Er is increasing (i) parallel to pr and (ii) anti-parallel to pr ? Sol. (a) Torque on an electric dipole in a uniform electric field d +q F+ E p q F+ FF–– –q dsinq node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxWe consider a dipole with charges +q and -q which are at a distance d away from each other. Let www.notesdrive.comit be placed in a uniform electric field of strength E such that the axis of the dipole forms an angle q with the electric field. The force on the charges is F+q = +qE ® towards the direction of electric field F–q = –qE ® opposite to the direction of electric field Since the magnitudes of forces are equal and they are separated by a distance d, The torque on the dipole is given by : Torque (t) = Force × perpendicular distance between both forces t = F × d sin q or t = qEdsinq So ttr==pprE´E.rsin[iqn (p = qd) or vector form] (b) If field is non uniform then dipole will experience both torque and force. (c) If external field is increasing parallel to pr then dipole will experience a force in the direction of external electric field and if field is increasing anti parallel to external field then dipole will experience a force against external field. 7. Define an equipotential surface. Draw equipotential surfaces : (i) In the case of a single point charge and (Refer to Ch-2, NCERT Q.7) (ii) In a constant electric field in Z-direction Why the equipotential surfaces about a single charge are not equidistant ? (iii) Can electric field exist tangential to an equipotential surface ? Give reason. Sol. (ii) Equipotential surface is a surface which has equal potential on every point of it. The net work done in moving a charge from one point to another on this surface is zero. A B E Z E

30 Physics ALLENÒ (iii) No, electric field cannot exist tangential to an equipotential surface. If it happens then a charged particle will experience a force along the tangential line and can move along it. Since a charged particle can move only due to the potential difference, this contradicts the concept of an equipotential surface. 8. Four point charges Q, q, Q and q are placed at the corners of a square of side 'a' as shown in the figure. Qq q aQ Find the (a) resultant electric force on a charge Q, and (b) potential energy of this system. OR (a) Three point charges q, –4q and 2q are placed at the vertices of an equilateral triangle ABC of side 'l' as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q. qA www.notesdrive.com node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx (b) –4q 2q Sol. (a) Bl C Find out the amount of the work done to separate the charges at infinite distance. Let us find the force on the charge Q at the point C. Force due to the other charge Q 1 Q2 1 æ Q2 ö =( )F14p Î0 = ç ÷ (along AC) 2 4p Î0 è 2a 2 ø a 2 Q q A B a2 CQ F3 F1 D qa F23 F2 Force due to the charge q placed at B F2 = 1 qQ along BC 4p Î0 a 2 E

ALLENÒ CBSE 31 Force due to the charge q placed at D F3 = 1 qQ along DC 4p Î0 a 2 Resultant of these two equal forces F2 and F3 ( )F23 = qQ 2 1 a2 (along AC) 4p Î0 \\ Net force on charge Q (at point C) F = F1 + F23 = 1 Q éQ + 2q ù 4p Î0 a2 ëê 2 ûú This force is directed along AC (For the charge Q, at the point A, the force will have the same magnitude but will be node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx directed along CA) www.notesdrive.com (b) Potential energy of the system = 1 éê4 qQ + q2 + Q2 ù 4p Î0 ë a a2 a ú 2 û = 1 a éê4qQ + q2 + Q2 ù 4p Î0 ë 2 2 ú û OR (a) Force on charge q due to the charge 4 q F1 = 1 æ 4q2 ö , along AB q 4p Î0 ç l2 ÷ è ø Force on the charge q due to the charge 2 q F2 = 1 æ 2q2 ö , along CA 4p Î0 ç l2 ÷ è ø F1 and F2 are inclinded to each other at an angle of 120º Hence, resultant electric force on charge q F = F12 + F22 + 2F1F2 cos q = F12 + F22 + 2F1F2 cos120º = F12 + F22 - F1F2 = æ1 q2 ö 16 + 4 -8 ç l2 ÷ è 4pÎ0 ø = 1 æ2 3q2 ö 4p Î0 çèç l2 ÷ø÷ E

32 Physics ALLENÒ (b) Net P.E. of the system = 1 . q2 [-4 + 2 - 8] 4pÎ0 l = (-10) q2 4pÎ0 l \\ Work done = 10q2 l = 5q2 4pÎ0 2pÎ0 l 9. Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery. C1=10µF P C2=20µF www.notesdrive.com A • 50µF •B node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxC3=5µFR C4=10µF Sol. The above circuit is a balanced wheatstone bridge. So there is no flow of current in 50mF capacitor (between P & R), so this capacitor will not play any role in the ckt. \\ modified ckt is Þ C2 C1, C2 are in series C1 •B C3, C4 are in series 1 1 1 A• C4 C' = 10mF + 20mF C3 C¢ AB C¢¢ Þ C' = 20 mF 3 Now C'' = 1 + 1 Þ C'' = 10 mF 5mF 10mF 3 Equivalent capacitance of the ckt = C¢+ C¢¢ Ceq = 20 10 = 10mF 3+3 Now, charge drawn from battery V = 10 volt (given) Q = CeqV Þ Q = 10 × 10 × 10–6 = 10–4C E

ALLENÒ CBSE 33 10. Three identical capacitors each of capacitance 3µF are connected, in turn in series and in parallel combination to the common source of V volt. Find out the ratio of the energies stored in two configurations. 3mF 3mF 3mF Sol. V Net capacitance = 1 = 1 + 1 + 1 C' 3 3 3 3mF 3mF node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 3mF www.notesdrive.com V Net capacitance Þ C¢ = 1mF Q Q = C'V C¢¢ = 3 + 3 + 3 = 9mF \\ Q = (10–6 × V)C Q = C\"V = (9 × 10–6 × V)C Energy (U) = 1 × QV Þ U2 = æ 1 ´ 9 ´ 10 –6 V 2 ö J 2 çè 2 ø÷ U1 = æ 1 ´10–6 V2 ö J èç 2 ø÷ Now, ratio U1 = 1 U2 9 11. In the following arrangement of capacitors, the energy stored in the 6 µF capacitor is E. Find the value of the following : (i) Energy stored in 12 µF capacitor. (ii) Energy stored in 3 µF capacitor. (iii) Total energy drawn from the battery +V – 6µF 3µF 12µF E

34 Physics ALLENÒ Sol. (i) Let EMF of applied battery in V. Energy stored in 6µF capacitor = E (given) V +– C1=6mF V3 C3=3mF V1 V2 C2=12mF Energy stored in a capacitor is given by www.notesdrive.comU=1CV2 Þ V= 2U 2 C node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx Here V1 = V2 îíìVV12 Þ Potential difference across 6mF capacitor Þ Potential difference across 12mF capacitor 2E1 = 2E2 C1 C2 E1 = E (given) E2 = C2 ´ E1 Þ E2 = 12 × E Þ E2 = 2E C1 6 (ii) Energy stored in 12µF capacitor (E2) = 2E C2=6mF QQ C3 = 3mF C2=12mF Total energy of parallel combination of capacitors is the sum of energy of both capacitors. Þ E + 2E = 3E So Q2 = 3E Þ Q= 2Ceq ´ 3E 2C eq Same charge (Q) flow through 3µF capacitor. So energy stored by capacitor of 3 µF E3 = Q2 = 2Ceq. ´ 3E Þ E3 = 2 ´18´ 3E 2C3 3C3 3´3 E3 = 18E (iii) Total energy drawn from battery = E1 + E2 + E3 = E + 2E + 18E = 21E E

ALLENÒ CBSE 35 12. Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of Îr = 4. XY +– 15 V (i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 µF (ii) Calculate the potential difference between the plates of X and Y. (iii) Estimate the ratio of electrostatic energy stored in X and Y Sol. (i) Let us assume that capacitance of X capacitor = C mF node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx \\ Capacitance of Y capacitor = 4C mF (As C = KC) www.notesdrive.com equivalent capacitance of system = 4mF (given) So 1 + 1 = 1 Þ C = 5mF 4C C 4 \\ Capacitance of Y capacitor ® 4C = 20mF (ii) We assume potential of capacitor X is V1 and V2 for capacitor Y. Q Q1 = Q2 [Both capacitor are in series] C1V1 = C2V2 1 = V2 Þ V2 = V1 ....(1) 4 V1 4 .....(2) Also given that V1 + V2 = 15V By solving above eq (1) and (2), we get V1 = 12V V2 = 3V (iii) Ratio of energy stored E1 = C1V12 = C1 ´çæ V1 ö2 = 1 ´çæè 4 ö2 Þ E1 = 4 E2 C2V22 C2 è V2 ÷ 4 1 ÷ø E2 1 ø 13. Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric. S VA B E

36 Physics ALLENÒ Sol. When switch 'S' is closed, the initial energy (Ei) of the system, Ei = 1 CV2 + 1 CV2 = CV2 2 2 If 1 CV2 = U, then Ei = CV2 = 2U ...(1) 2 When switch 'S' is opened then capacitor 'A' gets constant supply of voltage (V) & on the other hand, charge 'Q' becomes constant in capacitor 'B'. \\ In the above situation if dielectric of strength 'K' is filled, we may write. Final energy of the system as - Ef = 1 (KC)V2 + 1 æ Q2 ö 2 2 ç KC ÷ www.notesdrive.com è ø or node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxEf = KU +U=UæK2 +1öQ12 CV2=1Q2=U K ç K ÷ 2 C è ø Now, Ei = 2U(K) = 2K Ef U (K 2 + 1) K2 +1 EXERCISE-IV FIVE MARK QUESTIONS : 1. (i) Use Gauss’ law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities ? (Refer to NCERT Q. No 8) (ii) Find the ratio of the potential differences that must be applied across the series and parallel combination of two capacitors C1 and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same. Sol. (ii) Let C1 = x and C2 = 2x equivalent capacitance in series combination CS = C1C2 = x ´ 2x = 2x \\ CS = 2x C1 + C2 x + 2x 3 3 equivalent capacitance in parallel combination CP = C1 + C2 = x + 2x = 3x Now given that energy stored in series combination = Energy stored in parallel combination 1 Cs V12 = 1 C P V22 Þ 1 ´ æ 2x ö V12 = 1 ´ 3x ´ V22 Þ V1 = 3 2 2 2 çè 3 ÷ø 2 V2 2 E

ALLENÒ CBSE 37 2. (a) Define electric flux. Is it a scalar or a vector quantity ? A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown in the figure. Use Gauss' law to obtain the expression for the electric flux through the square. q d/2 d d (b) If the point charge is now moved to a distance 'd' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx OR www.notesdrive.com (a) Use Gauss' law to derive the expression for the electric field (Er ) due to a straight uniformly charged infinite line of charge density l C/m. (Refer to NCERT Q. No 9) (b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge. (c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2 > r1). Sol. (a) Electric flux through a given surface is defined as the dot product of electric field and area vector over that surface. ur r Alternatively f = òs E . dS Also accept Electric flux, through a surface equals the surface integral of the electric field over a closed surface. It is a scalar quantity. Constructing a cube of side 'd' so that charge 'q' gets placed within this cube (Gaussian surface) According to Gauss's law the Electric flux f = Ch arg e enclosed q = e0 e0 This is the total flux through all the six faces of the cube. Hence electric flux through the square 1 ´ q = q 6 e0 6e0 (b) If the charge is moved to distance d and the side of the square is doubled even then the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before. OR E

38 Physics ALLENÒ (b) The required graph is as shown : |E| Or (c) Work done in moving a charge q with displacement 'dr' r uur dW = F . dr ur uur dW = qE . dr = qEdr cos 0 dW = l dr 2pe0r www.notesdrive.comq´ Work done in moving the given charge from r1 to r2 (r2 > r1) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx r2 r2 lqdr r1 2pe0r = dW = r1 ò òW W = lq [loger2 – loger1] 2pe0 W = lq êéloge r2 ù 2pe0 ë r1 ú û 3. (a) Derive an expression for the electric field E due to a dipole of length '2a' at a point distant r from the centre of the dipole on the axial line. (Refer to Ch-1, NCERT Q.8) (b) Draw a graph of E versus r for r >> a. (c) If this dipole were kept in a uniform external electric field E0, with help of diagram represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases. OR (a) Use Gauss's theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density s. (Refer to Ch-1, NCERT Q.8) (b) An infinitely large thin plane sheet has a uniform surface charge density +s. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. E Sol. (b) for (r >> a) Or E

ALLENÒ CBSE 39 (c) (i) Stable equilibrium - (q = 0°) –q +q E0 Torque(t) = pE sinq = pE × sin0° = 0 (Q sin0° = 0) (ii) Unstable equilibrium node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx +q –q (q = 180°) www.notesdrive.com E0 Torque (t) = pE sinq = pE sin180° = pE × 0 = 0 (Q sin180° = 0) OR +s + + ++ ++ + P (b) + + + + +++ V= W = r Er.drr q ò ¥ r W = qò (-Edr) ¥ r W = -qóõô s dr 2 Î0 ¥ W = s (¥ - r) 2 Î0 W=¥ E

40 Physics ALLENÒ 4. (a) Draw a sketch of equipotential surface due to a single charge (–q) , depicting the electric field lines due to the charge. (b) Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side ‘a’ as shown below. q aa –4q a +2q Sol. (a) d3 d2 Equipotential –q d1 surface (concentric d3 > d2 > www.notesdrive.com (b) Total electrostatic potential energy of system node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 5. (i) é q(-4q) (-4q)(2q) q(2q) ù 10q 2 (ii) ëê a a a úû a U = U12 + U23 + U31 = k + + = -k ´ Sol. (b) (c) \\ Work done to dissociate the system W = –U W = 1 ´ 10q2 4pe0 a If two similar large plates, each of area A having surface charge densities +s and –s are separated by a distance d in air, find the expression for (a) Field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case. (Refer to 3 marks, PYQ.3) (b) The potential difference between the plates. (c) The capacitance of the capacitor so formed Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density s. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ? The difference between the plate : V = Ed Putting the value of E from equation no (1) V= sd/e0 ........(2) Capacitance of capacitor formed by plane sheets: Q = CV Putting the value of V from the equation (2) sA = C× sd/e0 C = e0 A/d E

ALLENÒ CBSE 41 (ii) Potential on the surface of charged metallic sphere is given by V = kq/r V= k.4pr2 s /r So V µ r It means that sphere having large radius will be at higher potential and charge always flow from higher potential to lower potential. So charge will flow from metallic sphere of radius 2R to sphere of radius R. When charge density s is same 6. (a) Distinguish, with the help of a suitable diagram, the difference in the behaviour of a conductor and a dielectric placed in an external electric field. How does polarised dielectric modify the original external field. ? (b) A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docxthen disconnected from the battery. If the separation between the plates of the capacitor is www.notesdrive.comnow doubled, how will the following change ? (i) Charge stored by the capacitor (ii) Field strength between the plates. (iii) Energy stored by the capacitor Justify your answer in each case. Sol. (a) When a conductor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static situation is achieved, i.e. when the two fields cancel each other and the net electrostatic field in the conductor becomes zero. sfree E0 E0 E0 E sin free E0 + Ein= 0 Conductor In contrast to conductors, dielectrics are non-conducting substances, i.e. they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching molecules of the dielectric. The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced, that does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of dielectric. E0 E0 s Ein sp E0 E0 + Ein ¹ 0 Dielectric E

42 Physics ALLENÒ Both polar and non-polar dielectrics develop net dipole moment in presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by Pr for linear isotropic dielectrics. (b) After disconnecting battery & doubling the separation between two plates of capacitor :- (i) Charge on capacitor remains same (due to charge conservation) i.e. CV = C¢V¢ Þ CV = æ C ö V ' Þ V¢ = 2V çè 2 ÷ø Voltage will be double. (ii) Field strength between the plates. E' = V' = 2V d' 2d www.notesdrive.comE'=V=E d node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx Electric field between the plates remains same. (iii) Energy stored in capacitor when connected to battery U1 = q2/2C Now energy stored in capacitor after disconnection from battery U2 = q2 = q2 /2 = q2 C 2C ' 2´C \\ Energy stored in capacitor get doubled to it’s initial value. 7. When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However, these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero. So no work is done in moving the electric dipole in uniform electric field. However, some work is done in rotating the dipole against the torque acting on it. (i) The dipole moment of a dipole in a uniform external field Er is p. Then the torque t acting on the dipole is :- (B) tr = pr ´ Er (C) tr = 2(pr + Er) (D) rt = (pr +Er) (A) t = pr.Er Ans. (B) rt = pr ´ Er (ii) An electric dipole consists of two opposite charges, each of magnitude 1.0 µC separated by a distance of 2.0 cm. The dipole is placed in an external field of 105 NC–1. The maximum torque on the dipole is (B) l × 10–3 Nm (C) 2 × 10–3 Nm (D) 4× 10–3 Nm (A) 0.2 × 10–3 Nm Ans. (C) 2 × 10–3 Nm (iii) Torque on a dipole in uniform electric field is minimum when q is equal to (A) 0° (B) 90° (C) 180° (D) Both (A) and (C) Ans. (D) Both (A) and (C) E

ALLENÒ CBSE 43 (iv) When an electric dipole is held at an angle in a uniform electric field, the net force F and torque t on the dipole are (A) F = 0, t ¹ 0 (B) F ¹ 0, t ¹ 0 (C) F = 0, t = 0 (D) F ¹ 0, t = 0 Ans. (A) F = 0, t ¹ 0 8. A dielectric slab is a substance which does not allow the flow of charges through it but permits them to exert electrostatic forces on one another. When a dielectric slab is placed between the plates, the field Eo polarises the dielectric. This induces charge –QP on the upper surface and +QP on the lower surface of the dielectric. These induced charges set up a field EP inside the dielectric in the opposite direction of Eo as shown. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx www.notesdrive.com (i) In a parallel plate capacitor, the capacitance increases from 4µF to 80µF on introducing a dielectric medium between the plates. What is the dielectric constant of the medium? (A) 10 (B) 20 (C) 50 (D) 100 Ans. (B) 20 (ii) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in second case. (A) 8pF (B) 10pF (C) 80pF (D) 100 pF Ans. (C) 80pF (iii) A dielectric introduced between the plates of a parallel plate capacitor (A) Increases the electric field between the plates (B) Decreases the capacity of the capacitor (C) Increases the charge stored in the capacitor (D) Increases the capacity of the capacitor Ans. (D) Increases the capacity of the capacitor (iv) A parallel plate capacitor of capacitance 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2 pF. What is the value of x. (A) 2 (B) 4 (C) 6 (D) 8 Ans. (B) 4 E

44 Physics ALLENÒ EXERCISE - V (RACE) CHAPTER-1 : ELECTRIC CHARGES & FIELDS For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : In electrostatics, electric field lines can never be closed loops. Reason : The number of electric field lines orginating from or terminating on a charge is proportional to the magnitude of charge. www.notesdrive.com (i) a (ii) b (iii) c (iv) d node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 2. Assertion : Under electrostatic conditions net electric field inside a solid conductor will be zero. Reason : Under electrostatics conditions there will be no free electrons inside a conductor. (i) a (ii) b (iii) c (iv) d 3. Assertion : Gauss law shows diversion when inverse square law is not obeyed. Reason : Gauss law is a consequences of conservation of charge. (i) a (ii) b (iii) c (iv) d 4. Assertion : Electrostatic force between two charges is a non conservative force. Reason : Electric force between two charges proportional to the square of distance between the two. (i) a (ii) b (iii) c (iv) d 5. Assertion : Electrostatic field lines are perpendicular to surface of conductor. Reason : Surface of a conductor is equipotential. (i) a (ii) b (iii) c (iv) d 6. Define dipole moment of an electric dipole. 7. Write the SI Unit of electric flux. 8. If in Coulomb's law, there would be dependence on 1 in place of, 1 , then would Gauss law be r3 r2 true? 9. A charge q is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube ? 10. What is the electric flux through a cube of side 1 cm which enclosed an electric dipole ? 11. What kind of charges are produced on each, when (i) a glass rod is rubbed with silk and (ii) an ebonite rod is rubbed with wool ? 12. Given two point charges q1 and q2, such that q1q2 < 0. What is the nature of force between them? 13. A charged particle is free to move in an electric field. Will it always move along an electric line of force? 14. Suppose a Gaussian surface does not include any net charge. Does it necessarily mean that E is equal to zero for all points on the surface? 15. How does electric field at a point charge with distance r from an infinite thin sheet of charge? E