Question

Three concentric conducting spherical shell have radii $r,2r$ and $3r$ and charges $q_1,q_2,\text{and}{q}_3$ respectively. Innermost and outermost shells are earthed as shown n figure. Select the correct alternative(s).

• A. $q_1=-\frac{q_2}{4}$
• B. $\frac{q_3}{q_2}=-\frac{1}{3}$
• C. $\frac{q_3}{q_1}=3$
• D. $q_1+q_3=-q_2$

Answer 1

Answer: (A), (C), (D)

$q_1+q_3=-q_2$

As the innermost and outermost spheres are earthed, so potential on these surfaces are zero.

Potential on innermost sphere is
$\frac{q_1}{r}+\frac{q_2}{2r}+\frac{q_3}{3r}=0$

$\Rightarrow 6q_1+3q_2+2q_3=0.......\left(1\right)$

Similarly, potential on outermost shell is
$\frac{q_1}{3r}+\frac{q_2}{3r}+\frac{q_3}{3r}=0$

$\therefore q_1+q_3=-q_2.......\left(2\right)$

Solving equation (1) and (2) we get
$6q_1+3q_2-2(q_1+q_2)=0$

$\therefore q_1=-\frac{q_2}{4}$

$6q_1-3(q_1+q_3)+2q_3=0$

$\therefore \frac{q_1}{q_3}=\frac{1}{3}$

and
$-6(q_2+q_3)+3q_2+2q_3=0$

$\therefore \frac{q_3}{q_2}=-\frac{3}{4}$

Thus, Option (a), (b) and (c) are correct.