Electrostatics: Two Questions (MCQ) on Sharing of Charge

(1) Three identical conducting spheres X, Y and Z carrying charges 8 μC, – 4.8 μC and 6.4 μC respectively are kept in contact and then separated from one another. Then, the sphere Y will have approximately

(a) an excess of 3×10¹³ electrons

(b) a deficiency of 3×10¹³ electrons

(c) a deficiency of 6×10¹³ electrons

(d) an excess of 6×10¹³ electrons

(e) a deficiency of 2×10¹³ electrons

The total charge on the spheres is (8 – 4.8 + 6.4) μC = 9.6 μC. Since the spheres are identical, they have the same capacitance and the charge is shared equally by them. Therefore, the charge on each sphere is 3.2 μC.

Since the charge on the sphere Y (and the spheres X and Z) is positive, there will be a deficiency of electrons.

Remembering that the electronic charge is 1.6×10^–19 coulomb, the deficiency of electrons on the sphere is 3.2×10^–6/1.6×10^–19 = 2×10¹³. Therefore the correct option is (e).

(2) Two capacitors C₁ and C₂ of capacitance 2 μF and 3 μF are charged to 50 volt and 40 volt respectively and arranged as shown, with the key K open. On closing the key, the charge flowing through the key will be

(a) 8 μC

(b) 12 μC

(c) 32 μC

(d) 64 μC

(e) 132 μC

You must remember that the total charge is conserved in all situations. In the above question, the total charge is C₁V₁ + C₂V₂ = 100 μC + 120 μC = 220 μC.

When the key K is closed, the capacitors get connected in parallel and the common potential difference between their terminals will be

V = (C₁V₁ + C₂V₂)/(C₁ + C₂) = 220 μC/ 5 μF = 44 volt.

The charge on C₁ after closing the key will be C₁V = 88 μC.

Since the initial charge on C₁ is 100 μC, the charge flowing through the key must be (100–88) μC = 12 μC.

[You will obtain the same result on considering C₂].