Title 1. Electric Charges and Fields.pdf ✅

Electric Charges and Coulomb's Law 1. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0μCm−2 . The charge on the sphere is nearly [AP/May 19, 2023 (I)] (a) 2.5 × 10−3C (b) 1.45 × 10−3C (c) 6.5 × 10−3C (d) 0.15 × 10−3C Ans: (b) Diameter of sphere d = 2.4 m Radius, r = d 2 = 1.2 m Surface charge density, σ = 80 × 10−6Cm−2 Charge on the sphere, Q = σA = σ(4πr 2 ) = 80 × 10−6 × 3.14 × (1.2) 2 = 1.45 × 10−3 2. Among the following the charge that does not exist on any type of charged body is [AP/May 18, 2023 (II)] (a) 3.2 × 10−19C (b) 6.4 × 10−19C (c) 9.6 × 10−20C (d) 9.6 × 10−18C Ans: (c) Charge on any body must be an integral multiple of charge on an electron or proton, i.e., charge is quantized, Q = ne If Q = 9.6 × 10−20 Then, n = Q e = 9.6×10−20 1.6×10−19 = 0.6 It is not a number, thus this charge does not exists. 3. An electron of charge ' e ' is moving round the nucleus of a hydrogen atom in a circular orbit of radius ' r '. The coulomb force F⃗ be- tween the two is ( here K = 1 4πε0 ) [AP/May 17, 2023 (II)] (a) −K e 2 r 3 rˆ (b) K e 2 r 3 r (c) −K e 2 r 3 r (d) K e 2 r 2 r Ans: (c) Using Coulomb's law, the force between two charges is given by F⃗ = kq1q2 r 3 r Here, charge of an electron, q1 = −e Charge of the nucleus, q2 = e F⃗ = k(−e)e r 3 r = − ke2 r 3 r 4. A charge Q is to be divided between two ob- jects. The values of the charges on the ob- jects so that the electrostatic force between them will be maximum is [AP/July 5, 2022 (II)] (a) Q 2 , Q 2 (b) Q 3 , 2 3 Q (c) Q 4 , 3 4 Q (d) Q 5 , 4 5 Q Ans: (a) Let charge on one object be ' q ', then on other object it is ' Q - q ' So, Fe = Kq(Q−q) r 2 . For Fe to be maximum, dFe dq = 0 Now, dFe dq = 0 ⇒ d dq [qQ = q 2 ] = 0 ⇒ Q − 2q = 0 ⇒ q = Q/2 Then, (Q − q) = Q − Q 2 = Q 2 So, force will be maximum when magnitude of charges are Q 2 and Q 2 . 5. A solid sphere of radius R carries a positive charge Q distributed uniformly throughout
its volume. A very thin hole is drilled through it's center. A particle of mass m and charge - q performs simple harmonic motion about the center of the sphere in this hole. The frequency of oscillation is [AP/July 4, 2022 (II)] (a) 1 2π [ Qq 4πε0R3 m ] 1 2 (b) 1 2π [ Qq 4πε0R2 m ] 1 2 (c) 1 2π Q [4πε0mR3] 1 2 (d) 1 2π [ Qq 4πε0mR] 1 2 Ans: (a) Force on charge ' q ' at distance ' r ' from centre is given by F⃗ = − 1 4πεo qQr⃗ R3 Clearly, F⃗ ∝ −r . so this is a SHM Comparing (i) with F = mω 2 r, we get mω 2 = 1 4πε0 qQ R3 ⇒ ω = √ 1 4πε0mR3 ⇒ f = 1 2π [ Qq 4πε0mR3 ] 1/2 6. An electron is released from a distance of 4 m from a stationary point charge 20nC. What will be the speed of the electron when it is 2 m away from the point charge? [Charge of electron = 1.6 × 10−19C, mass of elec- tron = 9 × 10−31 kg, 1 4πε = 9 × 109 S.I unit] [TS/July 18, 2022 (I)] (a) 2 × 106 m/s (b) 4 × 106 m/s (c) 1.6 × 106 m/s (d) 2.4 × 106 m/s Ans: (b) By law of conservation of mechanical en- ergy 0 − 4 4πε0 20 × 10−9 × 1.6 × 10−19 4 = −1 4πε0 20 × 10−9 × 1.6 × 10−19 2 + 1 2 mV2 ⇒ 9 × 109 × 32 × 10−28 4 + 9 × 109 × 32 × 10−28 2 = 1 2 × 9.1 × 10−31 V 2 ⇒ −7.2 × 10−18 + 14.4 × 10−18 = 4.55 × 10−31 V 2 ⇒ −7.2 × 10−18 = 4.55 × 10−31 V 2 ⇒ V = 3.977 × 106 ⇒ V = 4 × 106 m/s 7. Two charges are +10μC and −10μC are separated by 10 cm. The magnitude of force acting on another charge 5μC placed at the midpoint of the line joining the two charges will be [Use 1 4πε0 = 9 × 109 in SI unit] [TS/July 20, 2022 (I)] (a) 360 N (b) 0 N (c) 320 N (d) 380 N Ans: (a) F = F1 + F2 ⇒ F = 9 × 109 × 10 × 5 × 10−12 (5 × 10−2) 2 × 2 ⇒ F = 9 × 10 × 5 × 10−3 25 × 10−4 × 2 ⇒ F = 360 N
8. A charged cork ball having mass 1 g and charge q is suspended on a light string in a uniform electric field as shown in figure. The ball is in equilibrium at θ = 37∘ , when value of electric field is E = (3iˆ + 5jˆ) × 105NC−1 . (Assume, T as tension in the string.) Which of the following options are correct? (Given, sin 37∘ = 0.60 and g = 10 ms−2 ) [AP/Sept. 21, 2020 (I)] (a) q = 11 × 10−8C (b) T = 5.55 × 10−3 N (c) q = 12 × 10−9C (d) T = 4.55 × 10−3 N Ans: (b) Mass of cork ball, m = 1 g = 10−3 kg E̅ = (3iˆ + 5ˆj) × 105NC−1 Force on cork ball due to electric field E F = qE According to given diagram, equating forces in horizantal & vertical given direction, Tsin θ = qEx Tcos θ + qEy = mg ⇒ Tcos θ = mg + qEy Dividing Eqs. (i) by (ii), we get Tsin θ Tcos θ = qEx mg − qEy tan θ = qEx mg − qEy Putting values of known quantites tan 37∘ = q × 3 × 105 10−3 × 10 − q × 5 × 105 [As θ = 37∘ ] 3 4 = 3q × 105 10−2 − 5q × 105 [As tan 37∘ = 3/4 ] ⇒ 3 × 10−2 − 15q × 105 = 4 × 3q × 105 0.03 − 15 × 105q = 12 × 105q ⇒ q = 11 × 10−8C Putting value of q in Eq. (i). we get Tsin 37∘ = 1.1 × 10−8 × 3 × 105 T × 0.6 = 3.3 × 10−3 T = 3.3 0.6 × 10−3 = 5.5 × 10−3 N 9. Three charges 4q,Q and q are placed at po- sitions 0, l 2 and l respectively along a straight line. If the resultant force on q is zero, then Q is equal to [AP/Sept. 18, 2020 (I)] (a) −q (b) −2q (c) − q 2 (d) 4qt Ans: (a)
Resultant force on charge q is zero ∴ K ⋅ 4q ⋅ q l 2 + KQq (1/2) 2 = 0 ⇒ 4q l 2 + 4Q l 2 = 0 ⇒ q + Q = 0 ∴ Q = −q 10. Three charges of each magnitude 100μC are placed at the corners A, B and C of an equi- lateral triangle of side 4 m. If the charges at points A and C are positive and the charge at point B is negative, then the magnitude of total force acting on the charge at C and an- gle made by it with AC are [AP/Apr. 20, 2019 (I)] (a) 5.625 N, 60∘ (b) 0.5625 N, 60∘ (c) 5.625 N, 30∘ (d) 0.5625 N, 30∘ Ans: (a) Q = 100μC FA is force on charge at C due to A and FB is force on charge at C dec. to B. |FA | = |FB| = KQ 2 r 2 |Fnet | = |FA | = |FB| As angle between FA and FB is 120∘ . Putting the given values in Fnet , we get = 9 × 109 × (100 × 10−6 ) 2 (4) 2 Fnet = 5.625 N, 60∘ 11. Two small conducting balls of identical mass 20 g and identical charge 10−10C hang from non-conducting threads of length, L = 300 cm. If the equilibrium separation of balls is x and x ≪ L then the magnitude of x is (Assume, 4πε0 = 1 9×109 F/m and g = 10 m/s 2 ) [TS/May 3, 2019 (I)] (a) 2 5 1/3 mm (b) 3 101/3 mm (c) 3 1/3 10 mm (d) 3 2/3 5 mm Ans: (b) At point B, Tcos θ = mg Tsin θ = F Tsin θ = 1 4πε0 ⋅ q 2 x 2 From Eqn. (i) and (ii), we get Tsin θ Tcos θ = 1 4πε0 ⋅ q 2 x 2 mg ∴ mgtan θ = 9 × 109 ⋅ q 2 x 2 . From △ ABM, tan θ = x 2 √L 2 − ( x 2 ) 2 = x √4L 2 − x 2 = x 2L (∵ L ≫ x)