JEE Main 2026 Apr 6 Shift 1, Physics Q24: Electric Potential

JEE Main2026Apr 6, Shift 1Physics

A three coulomb charge moves from the point $(0, - 2, - 5)$ to the point $(5, 1, 2)$ in an electric field expressed as $E = 2 x \hat{i} + 3 y^{2} \hat{j} + 4 \hat{k}$ N/C. The work done in moving the charge is ________ J.

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Solution

Since $E$ is conservative (each component depends only on its own coordinate), the potential $V$ satisfies $E = - \nabla V$ , so $V (x, y, z) = - (x^{2} + y^{3} + 4 z) + C$ .

Potential difference between final point $(5, 1, 2)$ and initial point $(0, - 2, - 5)$ :

$Δ V = V_{f} - V_{i} = - [(25 + 1 + 8) - (0 - 8 - 20)] = - [34 - (- 28)] = - 62$ V.

Work done by external agent (or by the field; here taking the field's work since no other forces are specified): $W = - q Δ V$ ... here the work done by the electric force is $W_{elec} = q (V_{i} - V_{f}) = - q Δ V = - 3 (- 62) = 186$ J.

$∴ W = 186$ J.

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JEE Main 2026 Apr 6 Shift 1, Physics Q24: Electric Potential

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