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NUMERICALS-ANSWERS QUESTIONS ε0 = 8.85 × 10–12 C2 N– m–2 = × 109 Nm2 C–2 4πε0 g = 9.8 ms–2 e = 1.6 × 10–19 C Mass of electron me = 9.0 × 10–31 kg Mass of proton mp = 1.7 × 10–27 kg Charge of α particle = 3.2 × 10–19C Four point charges + µC + µC, + 10 µC and + µC are placed at the corners of a square of side 10 cm A charge q = + µC is placed at its centre find the net force on q Calculate the distance between two protons such that the electrical force between them is equal to the weight of either (AISSCE Delhi 1994) Two point charges are 0.1 m apart and their combined charge is µC If they repel each other with a force of 18 N, then calculate the magnitude of each charge Calculate the coulomb force between two alpha (α) particles separated by a distance of 3.2 × 10–15 m (AISSCE 1992) A proton moves through a uniform electric field of 5.01 × 103 NC–1 Calculate (a) the acceleration with which the proton is moving, and (b) the time taken by the proton to cover a distance of 4.8 cm Given mp = 1.67 × 10–27 kg S Chand & Company Limited What charge is required to produce a surface charge density of µC m–2 on a sphere of radius 77 cm Determine the ratio of the electrostatic and gravitational forces between a pair of electrons, placed in vacuum at a distance r from each other Charges of + 0.01 µC and – 0.01µC are placed at the vertices A & B, respectively, of an equilateral triangle ABC of side 0.3 m Calculate the field intensity at C Two point charges of magnitudes × 10–7 C and × 10–8 C are 12 cm apart (a) What electric field does each produce at the site of the other ? (b) What force acts on each ? 10 ABC is an equilateral triangle of side 10 m D is the mid-point of BC Charge + 100 µC, – 100 µC and + 75 µC are placed at B, C and D respectively What is the force experienced by a C positive charge placed at A ? 11 Two charges, each of + 16 µC, are placed along a line at a distance r apart A third charge –Q is placed between them Determine the position and value of Q so that the system is in equilibrium 12 Calculate the field due to an electric dipole of length 15 cm, and consisting of charges ± 50 µC, at a point 25 cm from each charge + 13 If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecule ion ( H ) In the ground state of H 2+ , the two protons are separated by 1.5 Å and the electron is Å from each proton Determine the potential energy of the system 14 A capacitor of capacitance 50µF is connected to a battery of 500 V Find the charge on it in coulomb and its energy in joule 15 A parallel plate capacitor consists of two circular plates of diameter 14 cm each Calculate the separation between the plates to have the capacitance equal to that of a sphere of radius 10 cm 16 Three capacitors of 3µF each are connected, first to have maximum capacitance and then to have minimum capacitance Find the ratio of the maximum to the minimum capacitance S Chand & Company Limited 17 Obtain the equivalent capacitance of the following network, given C1 = C4 = 100 pF and C2 = C3 = 200 pF Determine the charge and voltage across each capacitor (AISSCE 1990 C) 18 Show that the plates of a parallel plate capacitor attract each other with a force given by E D C1 F C C2 C3 B F = Q / 2ε0 A 300 V A C4 The symbols have their usual meaning 19 Calculate the electric potential on the surface of an isolated silver nucleus (Z = 47) having a radius of 3.4 × 10 –14 m 20 An oil drop of mass × 10–6 kg remain suspended at rest in a uniform vertical electric field Find the intensity of the electric field if the charge on the oil drop is equal to that of electrons 21 An electron is accelerated through a potential difference of 400 volts Calculate its kinetic energy and speed 22 How will you combine three capacitors, each of µF, so that their equivalent capacitance is (i) µF (ii) µF (iii) 4.5 µF? 23 A capacitor of 15 µF is charged to 500V and is then connected in parallel with another capacitor of µF charged to 200 V Find the common potential across the combination 24 Charge of –20 µC, 30 µC, 40 µC, and – 50 µC are placed at the four corners of a square of side 10 m Calculate the potential at the centre of the square 25 A positive charge of 10 C is placed at the origin of the coordinate system Calculate the work done in bringing a charge of 10 µC from infinity to the point P (0, 0, 18 cm) S Chand & Company Limited ANSWERS In the figure, F1, F2, F3 and F4 are the forces on the charge at O due to the charges at A, B, C and D respectively The forces F2 and F4 cancel each other Force at O due to + µC at A is F1 = q1 q2 4π ε r = = × 10 × × 10 −6 (5 2) × 10 9×5 50 × 10 r = OA = OC   = AP + OP   = 25 + 25 cm  = 50 cm  −6 −4 × 10 = N Similarly F3 = 18 N F1 and F3 are oppositely directed so the net force F = F3 – F1 = 18 – = 9N Let the required distance be r Then q1 q2 4πε0 r or r = = mg q1 q2 4πε0 mg S Chand & Company Limited C F1 F2 10 cm 10µ C D 2µ C 1µ C O F4 F3 A 5µ C P 2µ C B Here q1 = q2 = 1.6 × 10–19 C, m = 1.7 × 10–27 kg g = 9.8 m/s2 ∴r = 9 × 10 × 1.6 × 10 1.7 × 10 −19 −27 × 1.6 × 10 −19 × 9.8 r = 12.1 × 10–2 m = 12.1 cm Let one of the charges be q µC Then the other charge is (9–q) µC r = 0.1 m F = 18 N Solving we get, From Coulomb’s law, F = q1 q2 4πε0 r , we have 18 = or × 10 q (9 − q ) × 10 q (9 − q ) = (0.1) −6 × 10 −6 18 × 10 q – 9q + 20 = Solving, we get q = µC or µC Thus the two charges are µC and µC S Chand & Company Limited Coulomb’s force F = q1 q2 4πε0 r Here q1 = q2 = 3.2 × 10–19 C r = 3.2 × 10–15 m ∴F= × 10 × (3.2 × 10 −19 (3.2 × 10 ) (3.2 × 10 −19 ) −15 ) = 90 N (a) Force on the proton F = qE From Newton’s second law of motion, F = ma Therefore qE = ma a= or qE m Here q = e = 1.6 × 10–19 C E = 5.01 × 103 NC–1 and m = 1.67 × 10–27 kg ∴ a= 1.6 × 10 −19 × 5.01 × 103 1.67 × 10 −27 = 4.8 × 10 11 ms -2 S Chand & Company Limited (b) Using the equation s = ut + = at , we have t = 2s a × 4.8 × 10 (∵ u = 0) −2 11 4.8 × 10 -7 = 4.47 × 10 s Surface charge density σ = or q πr q = 4π r2σ Ηere r = 77 cm = 0.77 m and σ = µC/m2 = 10–6 Cm–2 So, q = 4× 22 × (0.77) × 10 −6 = × 22 × 77 × 11 × 10–10 = 7.45 ì 106 C = 7.45 àC Electrostatic force acting between two electrons Fe = ee πε r S Chand & Company Limited .(1) Gravitational force acting between two electrons for the same separation Fg = G me me r .(2) Dividing Eq (1) by Eq (2) Fe Fg Fe or Fg = e 2 4πε Gme = × 10 × (1.6 × 10 6.67 × 10 −11 −19 × (9 × 10 ) −31 ) = 4.3 × 1042 Electric field intensity E = E1 q C 4π ε r Field intensity at C due to the charge at A is E1 = × 10 × 10 60° −8 0.3 cm 0.3 × 0.3 = 1000 N/C (directed from A to C) Similarly, intensity at C due to the charge at B is 60° A + 0.01µ C E2 = 1000N/C (directed from C to B) S Chand & Company Limited E E2 B – 0.01µ C We note that E1 = E2 Resultant field at C = 2E1 cos 60° = × 1000 × (1/2) = 103 NC–1 parallel to AB Let q1 = × 10–7 C and q2 = × 10–8 C (a) Electric field due to q1 at the site of q2 is E1 = q1 4πε0 r = × 10 × × 10 (0.12) −7 = 1.875 × 105 NC–1 = 1.9 × 105 NC–1 Electric field due to q2 at the site of q1 is E2 = × 10 × × 10 (0.12) −8 = 5.625 × 104 NC–1 = 5.6 × 104 NC–1 (b) Force on each charge, F = q2 E1 = q1E2 = × 10–7 × 5.625 × 104 N = 1.7 × 10–2 N S Chand & Company Limited 10 Force acting on + 1C charge at A due to + 100 µC charge at B is FD × 109 × 100 × 10 −6 × FB = 100 A = × 103 N Force on the charge at A due to – 100 µC charge at C is −6 = × 103 N Force on the charge at A due to 75 àC charge at D is ì 09 × × −6 × ( 75 )2 = × 103 N F′ Fc ×1 10 × 10 FD = +1 C 10 m 9 × 10 × 100 × 10 FC = FB B + 100µ C  [ A D = A B − B D  2  = (1 ) − ( )  = 75 m  C D + 75µ C ] Resultant of FB and FC is ° F ′ = 2FB cos 60 = × 10 N FD and F ′ are mutually perpendicular So the net force on 1C charge at A is S Chand & Company Limited – 100µ C F = FD + F ′ 3 = (9 × 10 ) + (9 × 10 ) = ×10 N + 16µ C –Q + 16µ C 11 For the system to be in equilibrium, net force A O B on each charge must be zero Let charge –Q be placed at O at a distance x from the x r–x charge at A For charge –Q to be in equilibrium, the force on it due to the charge at A and that due to the charge at B must be equal in magnitude That is 4πε 16 × 10 x −6 Q ⇒ = x 16 × 10 4πε = −6 ( r − x) Q (r − x) or x = r/2 For the charge at A to be in equilibrium, the force on it due to the charge (–Q) and the force on it due to the charge at B must be equal and opposite So we have S Chand & Company Limited or −6 (16 × 10 ) Q πε x (16 × 10 ) Q 4πε ( r / 2) = −6 πε0 −6 = 16 × 10 × 16 × 10 r −6 −6 (16 × 10 ) 4πε r Q = × 10–6 C or The system will be in equilibrium when Q = – × 10–6 C and it is placed at the mid point of the line joining the two charges 12 The given point is on the equatorial line of the dipole, where the electric field is given by E= p C 3/ πε ( r + a ) Here p = 2aq 25 cm = (0.15) × (50 × 10–6) = 7.5 × 10–6 C-m 2 1/2 (r + a ) r = 25 cm = 0.25 m ∴ E= × 10 × 7.5 × 10 (0.25) −6 a A B 50à C + 50à C = 4.3 ì 106 N/C S Chand & Company Limited 15 cm 13 The total potential energy of the system –e B U = UAB + UBC + UCA U AB = U BC = U AC = ⇒ 4πε 10 −10 4πε 10 (e) (e) = = −10 4πε 1.5 × 10 −10 U= A  −e    4πε  10 −10   e  =  −10  4πε  1.5 × 10  ( − e ) ( + e) 2  − e2  e e − +  −10  −10 − 10 4πε 10 1.5 × 10 10  1 e 4πε 10 =− =− 4 −10   −1 − + 1.5  × × 10 × (1.6 × 10 10 × × 10 × (1.6 × 10 10 −10 −19 ) −10 × 1.6 × 10 J −19 ) −19 eV 1Å° 1Å°  − e2  =   4πε  10 −10  ( + e) ( − e) 1eV = 1.6 × 10−19 J  = − 19.2 eV S Chand & Company Limited +e 1.5 Å° Fig 1.34 C +e 14 C = 50 àF = 50 ì 106 F V = 500 V Q = CV = 50 × 10–6 × 500 = 25000 × 10–6 = 0.25 C Energy stored U= CV −6 = × 50 × 10 × 500 × 500 = 6.25 J 15 Let the separation between the plates be d The capacitance of a parallel plate capacitor is C = ε0 A d Capacitance of a spherical conductor is πε0r It is required that ε0 A d ⇒ d= = 4πε r A 4πr −2 = π (7 × 10 ) 4π × (0.1) –2 = 1.2 × 10 m = 1.2 cm S Chand & Company Limited 16 For maximum capacitance, the capacitors are connected in parallel Cmax = C1 + C2 + C3 = + + = µF For minimum capacitance, the capacitors are connected in series 1/C = 1/3 + 1/3 + 1/3 = or 1 C = 1µF C max C = =9 17 The network may be redrawn as in Fig 1.35 C1 = C4 = 100 pF C2 = C3 = 200 pF C2 and C3 are in series So the equivalent capacitance between C and F is C′ = 200 = 100 pF C ′ is in parallel with C1 So the capacitance between B and G is C′′ = 100 + 100 = 200 pF Equivalent Capacitance between A and G is S Chand & Company Limited C= = C4 C ′′ C4 + C ′′ 100 × 200 300 = 66.7 pF The potential differences between A and B and between B and G will be in the inverse ratio of their capacitances So potential difference between A and B = 200 V Potential difference between B and G = 100 V Potential difference between D and E = 100 V Potential difference across C2 or across C3 will be equal to 50 V Q1 = C1 V1 = 10–8 C Q3 = Q2 = C2 V2 = C3 V3 = 10–8 C Q4 = C4 V4 = × 10–8 C 18 Energy stored in the capacitor with plate separation x is U= Q 2C = Q 2 (ε A / x) = Q x 2ε0 A If the plate separation is increased to x + dx, the stored energy becomes U= Q ( x + dx) 2ε0 A S Chand & Company Limited Therefore, the work done to increase the plate separation by dx is dW = U ′ − U = Q dx 2ε A If F is the force between the plates, then we also have dW = F dx Eqs (1) and (2) give .(1) .(2) F = Q2/2ε0A Note : This result can also be expressed in terms of the electric field between the plates : F= Q Q  σ ε = Q ε = QE  0A 2 19 We have Q = Ze = 47 × 1.6 × 10–19 C = 7.52 × 10–18 C Now V = Q 4πε0 r = × 10 × 7.52 × 10 3.4 × 10 −18 −14 = 2.0 ×10 V S Chand & Company Limited 20 The direction of the field must be downward so that the electrical force on the drop is upward We have QE = mg where Q = ne Therefore, E= mg ne –6 Here m = × 10 kg, g = 9.8 m/s2 , n = 8, e = 1.6 × 10–19 C So, E= × 10 −6 × 9.8 × 1.6 × 10 −19 13 = 3.1 × 10 NC −1 21 Kinetic Energy acquired by a charge q accelerated through a potential differences V is qV Here, q = 1.6 × 10–19 C; V = 400 V So, Ek = 1.6 × 10–19 × 400 = 6.4 × 10–17 J If v is the speed of the electron, then Ek = or mv v = Ek / m = × 6.4 × 10 × 10 −17 −31 = 1.2 ×10 ms –1 S Chand & Company Limited 22 (i) In parallel C = C1 + C2 + C3 = + + = µF (ii ) In series C = C1 + C2 + C3 ⇒ C = + + =1 or C = 1µF (iii) Two capacitors should be connected in series and the third in parallel with the series combination (Fig 1.37) C= = C1 C2 C1 + C2 3ì + C3 C1 3àF 3àF +3 C3 3àF = 1.5 + = 4.5 µF Fig 1.36 C2 23 If the charge on the first capacitor is Q1and that on the second capacitor is Q2, then Q1 = C1 V1 = 15 × 10–6 × 500 = 75 × 10–4 C Q2 = C2 V2 = × 10–6 × 200 = × 10–3 C = 10 × 10–4 C S Chand & Company Limited Common potential V = = Q1 + Q2 C1 + C (75 × 10−4 ) + (10 × 10−4 ) (15 × 10−6 ) + (5 × 10−6 ) = 425 Volts 24 AO = BO = CO =DO = = = = 1 (diagonal AC or BD) AB + BC 2 D (10 ) + (10 ) – 50µ C 40à C C O ì 20 = 10 m Total potential at O A – 20µ C V = VA + VB + VC + VD VA = kQA r = VB = × 10 ( −20) 10 kQB r = × 10 × 30 10 = − 18 × 10 µV = 27 × 10 µV S Chand & Company Limited 30à C B Similarly VC = 36 ì 109 àV, VD = 45 ì 109 àV V = (– 18 + 27 + 36 – 45) × 109 V = Zero 25 Potential at the point P due to a charge Q at the origin is V= kQ r Here Q = 10 C and r = 18 cm = 0.18 m So, V= × 10 × 10 0.18 11 Work done = × 10 V W = Vq = × 1011 × 10 × 10–6 = × 106 J S Chand & Company Limited