MHTCET 16 Apr 2026 Shift 2 Paper

Concept:The electrostatic potential energy of $q_3$ due to $q_1$ and $q_2$ is $U = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_3}{r_1} + \frac{q_2 q_3}{r_2}\right)$.The change in energy depends only on how the distances $r_1$ and $r_2$ change when $q_3$ moves from C to D.Explanation:From the figure, $AB = 30 \text{ cm} = 0.3 \text{ m}$ and $AC = 40 \text{ cm} = 0.4 \text{ m}$.Triangle ABC is right-angled at A, so $BC = \sqrt{(0.3)^2 + (0.4)^2} = 0.5 \text{ m}$.Thus, potential energy at C is $U_C = \frac{q_3}{4\pi\varepsilon_0}\left(\frac{q_1}{0.4} + \frac{q_2}{0.5}\right)$.When $q_3$ moves along the circular path of radius 40 cm centered at A, $AD = AC = 0.4 \text{ m}$.Points A, B, and D are collinear, so $BD = AD - AB = 0.4 - 0.3 = 0.1 \text{ m}$.Thus, potential energy at D is $U_D = \frac{q_3}{4\pi\varepsilon_0}\left(\frac{q_1}{0.4} + \frac{q_2}{0.1}\right)$.The $q_1$ terms cancel because $AC = AD$.So, $\Delta U = U_D - U_C = \frac{q_3}{4\pi\varepsilon_0}\left(\frac{q_2}{0.1} - \frac{q_2}{0.5}\right)$.This gives $\Delta U = \frac{q_3 q_2}{4\pi\varepsilon_0}(10 - 2) = \frac{8q_2 q_3}{4\pi\varepsilon_0}$.Comparing with the given form $\frac{q_3 k}{4\pi\varepsilon_0}$, we get $k = 8q_2$.Answer:$k = 8q_2$, so option A is correct.