A point charge of 10⁻⁸ C is placed at origin. The work done in moving a point charge 2 from point A(4, 4, 2) m to B(2, 2, 1) m is ______ J. ((1)/(4πε₀) = 9 × 10⁹ in SI units) (a) 30 × 10⁻⁶ (b) 0 (c) 15 × 10⁻⁶ (d) 45 × 10⁻⁶

Answer: (a). rA = √(16+16+4) = 6 m, rB = √(4+4+1) = 3 m. VA = 9×10⁹ × 10⁻⁸6 = 15 V, VB = 90/3 = 30 V. W = q(VB - VA) = 2×10⁻⁶ × 15 = 30×10⁻⁶ J So the answer is (a) 30×10⁻⁶ J.

A point charge of 10⁻⁸ C is placed at origin. The work done in moving… | JEE Main 2026 (21 Jan, Shift 1)

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Electrostatic Potential Energy

A point charge of $10^{-8}\text{ C}$ is placed at origin. The work done in moving a point charge $2\ \mu\text{C}$ from point $A(4, 4, 2)\text{ m}$ to $B(2, 2, 1)\text{ m}$ is ______ J. $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$ in SI units)

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Answer: (a)

$r_A = \sqrt{16+16+4} = 6$ m, $r_B = \sqrt{4+4+1} = 3$ m.
$V_A = \dfrac{9\times10^9 \times 10^{-8}}{6} = 15$ V, $V_B = \dfrac{90}{3} = 30$ V.
$W = q(V_B - V_A) = 2\times10^{-6} \times 15 = 30\times10^{-6}$ J
So the answer is (a) $30\times10^{-6}$ J.

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