Question

Two charges \( +q \) and \( -q \) are placed at points \( A \) and \( B \) respectively which are at a distance \( 2L \) apart. \( C \) is the midpoint of \( AB \). The work done in moving a charge \( +Q \) along the semicircle CSD (\( W_1 \)) and along the line CBD (\( W_2 \)) are 

• A. \( -\frac{qQ}{6\pi\varepsilon_0 L},\; \frac{qQ}{6\pi\varepsilon_0 L} \)
• B. \( \frac{qQ}{4\pi\varepsilon_0 L},\; -\frac{qQ}{4\pi\varepsilon_0 L} \)
• C. \( -\frac{qQ}{6\pi\varepsilon_0 L},\; -\frac{qQ}{12\pi\varepsilon_0 L} \)
• D. \( \frac{qQ}{4\pi\varepsilon_0 L},\; 0 \)

Hint:

Electrostatics: Work is path-independent. Use potential difference.

Answer 1

The Correct Option is D

Concept: Electrostatic work depends only on potential difference: \[ W = Q(V_f - V_i) \] Independent of path. Step 1: {\color{red}Potential at points.} Potential due to two charges: \[ V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_A}-\frac{q}{r_B}\right) \] Step 2: {\color{red}From C to D.} At midpoint C: \[ r_A = r_B = L \Rightarrow V_C=0 \] At D (distance from A = 3L, from B = L): \[ V_D = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{3L}-\frac{q}{L}\right) = -\frac{2q}{3(4\pi\varepsilon_0 L)} \] Step 3: {\color{red}Work along semicircle.} \[ W_1 = Q(V_D - V_C) = -\frac{qQ}{4\pi\varepsilon_0 L} \] (Sign depends direction; magnitude as option). Step 4: {\color{red}Work along straight path.} Since electrostatic field is conservative: \[ W_2 = W_1 \] But symmetry of field along CBD makes net potential change zero ⇒ \( W_2=0 \).