Title 2.ELECTROSTATIC POTENTIAL AND CAPACITANCE - Explanations.pdf ✅

1 (d) Work done = U2 − U1 = q1q2 4πε0 [ 1 r2 − 1 r1 ] = 12 × 10−6 × 8 × 10−6 × 9 × 109 [ 102 6 − 102 10 ] W = 96 × 9 × 10−3 × 102 × 4 60 = 5.8J 3 (b) Potential at centre due to all charges are = 1 4πε0 [ q d + q d + q d + q d ] = 1 4πε0 4q d in SI units = 4q d in CGS units 4 (b) E = λ 2πε0r λ = 2πε0r E = 1 2 × 9 × 109 × 2 × 10−2 × 9 × 0 4 = 10−7 Cm−1 5 (a) Work done in delivering q coulomb of charge from clouds to ground. W = Vq = 4 × 106 × 4 = 16 × 106 J The power of lightning strike is P = W t = 16 × 106 0.1 = 160 × 106W = 160 MW 6 (d) C = ε0A d1 K1 + d2 K2
= ε0A d 2 ( 1 1 + 1 2 ) = 4ε0A 3d 7 (c) Net emf in the circuit here E = E2 − E1 = 16 − 6 = 10 volt While the equivalent capacity C = C1C2 C1+C2 = 2×3 2+3 = 6 5 μF Charge on each capacitor q = CV = 6 5 × 10 = 12μC ∴ Potential difference across 2μF capacitor V1 = q C1 = 12 2 = 6 volt 8 (c) For a ring E = 1 4πε0 . qx (x 2+R2) and E is maximum when x = R √2 ⇒ Emax = 1 4πε0 . 2q 3√3. R2 9 (a) The 10μF and 6μF capacitors are connected in parallel, hence resultant capacitance is C ′ = 10 μF + 6 μF = 16 μF This is connected in series with 4 μF capacitor, hence effective capacitance is 1 C′′ = 1 16 + 1 4 = 20 16 × 4 ⇒ C ′′ = 64 20 = 3.20μF 10 (a) ∵ V = 4x 2 Hence, E⃗ = − dV dr = −8xi̇ ̂ Hence, value of E⃗ at (1m, 0, 2m) will be E⃗ = −8 × 1i̇ ̂ = −8i̇ ̂ Vm−1 11 (a) The situation is summarised in figure. BC = AD = 3cm, AB = DC = 4cm, So, AC = 5cm.
Now, potential at A VA = 1 4πε0 qB AB + 1 4πε0 . qc AC + 1 4πε0 . qD AD = 1 4πε0 [ 10 × 10−12 4 × 10−2 − 20 × 10−12 5 × 10−2 + 10 × 10−12 3 × 10−2 ] = 9 × 109 × 10−10 [ 10 4 − 20 5 + 10 3 ] = 9 × 10−1 × 11 6 = 16.5 × 10−1 = 1.65V 12 (c) 6 μF and 3 μF capacitors are in series 1 C1 = 1 6 + 1 3 C1 = 2 C1 is parallel to 2 μF capacitor ∴ Ceq = 2 + 2 = 4μF Total energy, U = 1 2 CV 2 = 1 2 × 4 × (2) 2 = 8μJ 13 (c) Capacitance C = Q V For a dielectric media C = εA d ∴ Capacitance C of a capacitor is independent of the geometrical configuration of the capacitor. 14 (a) Potential gradient relates with electric field according to the relation,E = − dV dr = − 10 20 × 10−2 = 50 Vm−1 16 (d) Positive plate of all the three condensers is connected to one point (A) and negative plate of all the three condensers is connected to point (B) ie, they are joined in parallel. Cp = 3 + 3 + 3 = 9μ F 17 (c) The situation is shown in the figure. Plate 1 has surface charge density σ and plate 2 has surface charge density– σ. The electric fields
at point P due to two charged plates add up, giving E = σ 2ε0 + σ 2ε0 = σ ε0 Given , σ = 26.4 × 10−12Cm−2 ε0 = 8.85 × 10−12C 2N −1m−2 Hence , E = 26.4×10−12 8.85×10−12 ≈ 3NC−1 Note the direction of electric field is from the positive to the negative plate. 18 (c) On sharing of charges loss in electrical energy, ΔU = C1C2 2(C1+C2 ) (V1 − V2 ). In present case C1 = C2 = C ∴ ΔU = C 2 2(2C) (V1 − V2 ) 2 = 1 4 C(V1 − V2 ) 2 19 (a) The potential V = 1 4πε0 1 r (q1 + q2 + q3 + q4 ) V = (9 × 109) ( 1 1m ) {(3 − 3 − 4 + 7) × 10−6C} = 2.7 × 104 N − m C = 2.7 × 104 JC −1 = 2.7 × 104V 20 (d) (n-1) capacitors are made by n plates and all are connected in parallel because plates are connected alternately. ∴ Total capacitance = (n − 1)x 21 (a) Required work done, W = QV = (2e) × 25 = 50e = 50 × 1.6 × 10−19 = 8 × 10−18J 22 (c) As battery is disconnected, total charge Q is shared equally by two capacitors. energy of each capacitor + + + + + + + + + + + + + - - - - - - - - - - - - - - - P 1 2 - E