GATE 2026 EE Q.63: solution and answer

GATE 2026 · EEQ.63 · 2 marks · NAT

A uniform spherical volume charge distribution of radius 2 m, centered at the origin, has a strength of $\frac{3}{π} \times 1 0^{- 6} C/m^{3}$ . A point charge of strength $π \times 8.854 \times 1 0^{- 12} C$ is moved from $(- 3, 0, - 4)$ to $(0, 0, 4)$ in Cartesian coordinate system. The relative permittivity of the medium is 1 and the coordinate values are in meters.

The work done during the process is ______ $μ J$

(Round off to two decimal places)

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Answer and solution

ANSWER

The work done is 0.40 μJ.

Official answer: 0.38 to 0.42Source: Official answer key, GATE 2026 EE (IIT Guwahati)

Locate the end points. Start $A (- 3, 0, - 4)$ : $r_{A} = 9 + 0 + 16 = 5 m$ . End $B (0, 0, 4)$ : $r_{B} = 0 + 0 + 16 = 4 m$ . Both lie outside the charged sphere of radius $R = 2 m$ .
Total charge of the sphere. $Q = ρ_{v} \cdot \frac{4}{3} π R^{3} = \frac{3}{π} \times 1 0^{- 6} \times \frac{4}{3} π (2)^{3} = 32 \times 1 0^{- 6} C$
Potential outside the sphere. Gauss's law over a concentric sphere of radius $r \geq R$ gives $E = \frac{Q}{4 π ϵ _{0} r ^{2}} \overset{a}{^}_{r}$ , the same as a point charge $Q$ at the origin. Integrating from infinity, $V (r) = \frac{Q}{4 π ϵ _{0} r}$ for $r \geq R$ , with $ϵ_{r} = 1$ and $ϵ_{0} = 8.854 \times 1 0^{- 12} F/m$ .
Work done by the external agent is $W = q (V_{B} - V_{A})$ . The electrostatic field is conservative, so this depends only on the end points and not on the path: $W = \frac{q Q}{4 π ϵ _{0}} (\frac{1}{r _{B}} - \frac{1}{r _{A}})$
Use the given value of q. $q = π \times 8.854 \times 1 0^{- 12} C = π ϵ_{0}$ , so $\frac{q}{4 π ϵ _{0}} = \frac{π ϵ _{0}}{4 π ϵ _{0}} = \frac{1}{4}$ .
Compute. $W = \frac{32 \times 1 0 ^{- 6}}{4} (\frac{1}{4} - \frac{1}{5}) = 8 \times 1 0^{- 6} \times 0.05 = 4 \times 1 0^{- 7} J = 0.40 μ J$
Check the sign. The work is positive: the positive point charge is pushed from $r = 5 m$ to $r = 4 m$ , closer to the positively charged sphere, against the repulsion.

KEY IDEA

Outside a uniformly charged sphere, the field and potential are those of a point charge $Q$ at the centre. Electrostatic work depends only on the end points, $W = q (V_{B} - V_{A})$ , so only the two radial distances matter (5 m and 4 m). The point charge is given as $π ϵ_{0}$ so that $q / (4 π ϵ_{0}) = 1/4$ exactly.

WATCH OUT FOR

Three common mistakes: (1) using the $z$ -coordinates or the straight-line distance between the points, 5 m apart, instead of each point's distance from the origin; (2) using the interior potential formula when both points are outside the sphere ( $r_{A} = 5$ , $r_{B} = 4 > R = 2$ ); (3) writing $V_{A} - V_{B}$ , which gives $- 0.40 μ J$ (the work done by the field) instead of the work done by the external agent in moving the charge.

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