Q: A solid conducting sphere having a charge Q is surrounded by an uncharged conducting hollow spherical shell. Let the pot. diff. between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge – 3 Q , the new potential difference between the same two surface is

(a) V

(b) 2 V

(c) 4 V

(d) – 2 V

Ans: (a)

Sol: Potential of inner sphere = Q/R_{1}

Potential of outer surface of shell = Q/R_{2}

Potential difference , $ \displaystyle V_1 = \frac{Q}{R_1} -\frac{Q}{R_2} $

When a charge – 3 Q is given to the shell,

Potential of inner sphere , $\displaystyle = \frac{Q}{R_1} -\frac{3Q}{R_2} $

Potential of outer surface of shell , $\displaystyle = \frac{Q}{R_2} -\frac{3Q}{R_2} = -\frac{2Q}{R_2}$

Potential difference , $\displaystyle V_2 = \frac{Q}{R_1} -\frac{3Q}{R_2} – (-\frac{2Q}{R_2}) $

$ \displaystyle V_2 = \frac{Q}{R_1} -\frac{Q}{R_2} $

i.e. V_{2} = V_{1} = V