**(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****×10 ^{13} electrons**

**(b) a deficiency of 3****×10 ^{13} electrons**

**(c) a deficiency of 6****×10 ^{13} electrons**

**(d) an excess of 6****×10 ^{13} electrons**

**(e) a deficiency of 2****×10 ^{13} 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} = 2×10^{13}. Therefore the correct option is (e).

**(2) Two capacitors C _{1} and C_{2 }of capacitance 2 μF and 3 μF are charged to 50 volt and 40 volt respectively and arranged as shown, with the key K ope**

**n. 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 C_{1}V_{1} + C_{2}V_{2} = 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 = (C_{1}V_{1} + C_{2}V_{2})/(C_{1} + C_{2}) = 220 μC/ 5 μF = 44 volt.

The charge on C_{1} after closing the key will be C_{1}V = 88 μC.

Since the initial charge on C_{1} is 100 μC, the charge flowing through the key must be (100–88) μC = 12 μC.

[You will obtain the same result on considering C_{2}]._{}