Question

Figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B, is equal to :

• A. $-\frac{3Q}{4}$
• B. $-\frac{8Q}{11}$
• C. $-\frac{5Q}{3}$
• D. $-\frac{3Q}{7}$

Answer 1

The correct option is A $-\frac{3Q}{4}$

Concentric meter shells $=A,B,C$

radii $=a,2a,3a$ respectively.

Shell $B$ is earthier and shell $C$ is given by charge $Q$.

$V=\frac{Kq}{r}$

The attempt at a solution charge $q$ is supplied from earth to shell.

Now, we have to find $q$, Potential of shell $C$ should be zero.

$\frac{K(q^{{}^{\prime }}+q)}{3a}.\frac{Kq}{3a}+\frac{Kq}{a}.\frac{Kq}{a}+\frac{Kq}{2a}=0$

$\frac{K(q^{{}^{\prime }}+q)}{3a}\frac{Kq}{3a}+\frac{Kq}{2a}=0$

$\frac{K(q^{{}^{\prime }}+q)+Kq+\frac{3Kq}{2}}{3a}=0$

or, $2K(q^{{}^{\prime }}+q).2Kq+3Kq=0$

or, $2K(q^{{}^{\prime }}+q)+Kq=0$

or, $2K(q^{{}^{\prime }}+q)=-Kq$

or, $2(q^{{}^{\prime }}+q)=-q$

or, $q^{{}^{\prime }}+q=-\frac{q}{2}$

or, $q^{{}^{\prime }}=-\frac{q}{2}.q$

or, $q^{{}^{\prime }}=\frac{-3q}{2}$