Electrostatics Part 3

Concept:Work done by an external agent in moving a charge in an electrostatic field is equal to the change in electrostatic potential energy.Explanation:Given: source charge $Q = 10^{-8}\ \mathrm{C}$ at origin.Test charge $q = 2\ \mu\mathrm{C} = 2 \times 10^{-6}\ \mathrm{C}$.Distance of point $(x, y, z)$ from origin: $r = \sqrt{x^2 + y^2 + z^2}$.For point $A(4, 4, 2)$: $r_A = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\ \mathrm{m}$.For point $B(2, 2, 1)$: $r_B = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\ \mathrm{m}$.Potential due to point charge: $V = k \frac{Q}{r}$, where $k = 9 \times 10^{9}\ \mathrm{Nm^2/C^2}$.$V_A = \frac{(9 \times 10^9)(10^{-8})}{6} = \frac{90}{6} = 15\ \mathrm{V}$.$V_B = \frac{(9 \times 10^9)(10^{-8})}{3} = \frac{90}{3} = 30\ \mathrm{V}$.Work done: $W = q (V_B - V_A) = (2 \times 10^{-6})(30 - 15) = (2 \times 10^{-6})(15) = 30 \times 10^{-6}\ \mathrm{J}$.Answer:Work done is $30 \times 10^{-6}\ \mathrm{J}$, which corresponds to option (A).