GATE 2022 EE Q.53: solution and answer

GATE 2022 · EEQ.53 · 2 marks · MCQ

As shown in the figure below, two concentric conducting spherical shells, centered at $r = 0$ and having radii $r = c$ and $r = d$ are maintained at potentials such that the potential $V (r)$ at $r = c$ is $V_{1}$ and $V (r)$ at $r = d$ is $V_{2}$ . Assume that $V (r)$ depends only on $r$ , where $r$ is the radial distance. The expression for $V (r)$ in the region between $r = c$ and $r = d$ is

GATE 2022 EE question 53: Two concentric circles, both centred at the origin. — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
$V (r) = \frac{c d ( V _{2} - V _{1} )}{( d - c ) r} - \frac{V _{1} c + V _{2} d - 2 V _{1} d}{d - c}$
(B)
$V (r) = \frac{c d ( V _{1} - V _{2} )}{( d - c ) r} + \frac{V _{2} d - V _{1} c}{d - c}$
(C)
$V (r) = \frac{c d ( V _{1} - V _{2} )}{( d - c ) r} - \frac{V _{1} c - V _{2} c}{d - c}$
(D)
$V (r) = \frac{c d ( V _{2} - V _{1} )}{( d - c ) r} - \frac{V _{2} c - V _{1} c}{d - c}$

Answer and solution

ANSWER

Option (B): $V (r) = \frac{c d ( V _{1} - V _{2} )}{( d - c ) r} + \frac{V _{2} d - V _{1} c}{d - c}$ .

Official answer: BSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

The shells are conductors, so any charge sits on their surfaces and the region $c < r < d$ contains no charge. The potential there therefore obeys Laplace's equation, $\nabla^{2} V = 0$ . Because $V$ depends only on $r$ , in spherical coordinates this becomes $\frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d V}{d r}) = 0$
Integrate once: $r^{2} \frac{d V}{d r}$ is a constant. Call it $- A$ , so $\frac{d V}{d r} = - \frac{A}{r ^{2}}$ . Integrating again gives $V (r) = \frac{A}{r} + B$ where $A$ and $B$ are constants.
Apply the boundary conditions: $\frac{A}{c} + B = V_{1}$ and $\frac{A}{d} + B = V_{2}$ .
Subtract the second equation from the first: $A (\frac{1}{c} - \frac{1}{d}) = V_{1} - V_{2}$ . Since $\frac{1}{c} - \frac{1}{d} = \frac{d - c}{c d}$ , this gives $A = \frac{c d ( V _{1} - V _{2} )}{d - c}$
Back-substitute into the second condition: $B = V_{2} - \frac{A}{d} = V_{2} - \frac{c ( V _{1} - V _{2} )}{d - c} = \frac{V _{2} ( d - c ) - c V _{1} + c V _{2}}{d - c} = \frac{V _{2} d - V _{1} c}{d - c}$
So $V (r) = \frac{c d ( V _{1} - V _{2} )}{( d - c ) r} + \frac{V _{2} d - V _{1} c}{d - c}$ which is option (B).
Check option (B) at both walls. At $r = c$ : $\frac{d ( V _{1} - V _{2} ) + V _{2} d - V _{1} c}{d - c} = \frac{V _{1} ( d - c )}{d - c} = V_{1}$ . At $r = d$ : $\frac{c ( V _{1} - V _{2} ) + V _{2} d - V _{1} c}{d - c} = \frac{V _{2} ( d - c )}{d - c} = V_{2}$ . Both conditions hold.
The other options fail the same check. (A) does give $V_{1}$ at $r = c$ , but at $r = d$ it gives $\frac{c ( V _{2} - V _{1} ) - V _{1} c - V _{2} d + 2 V _{1} d}{d - c} = 2 V_{1} - V_{2}$ , not $V_{2}$ . (C) gives $\frac{( V _{1} - V _{2} ) ( d - c )}{d - c} = V_{1} - V_{2}$ at $r = c$ . (D) gives $V_{2} - V_{1}$ at $r = c$ . Neither of the last two equals $V_{1}$ in general.

KEY IDEA

No charge sits between the shells, so $V$ obeys Laplace's equation. With spherical symmetry the general solution is $V (r) = \frac{A}{r} + B$ , and the two shell potentials fix $A$ and $B$ . In an exam, the fastest route is to put $r = c$ and $r = d$ into each option. Only the correct expression gives $V_{1}$ and $V_{2}$ .

WATCH OUT FOR

Checking only one boundary. Option (A) also gives $V_{1}$ at $r = c$ , and fails only at $r = d$ , where it gives $2 V_{1} - V_{2}$ . Check both walls. Other slips are using the cylindrical solution $A ln r + B$ instead of the spherical $A / r + B$ , and losing a sign when solving for $B$ .

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 23 Electrostatics questions, with notes

GATE 2022 EE Q.53: solution and answer

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