Question

Two points A and B are located at distances 4 cm and 5 cm respectively from a point charge $+20\mu C$. The work done in taking a charge of $+2\mu C$ from point A to point B is

• A. 0.9 J
• B. 3.6 J
• C. 1.8 J
• D. 1.2 J

Hint:

$W_{ext} = \Delta U$. Repulsion moves charges apart naturally, so potential energy decreases. $W_{ext}$ is negative, meaning energy is output, not input. Pick the magnitude match.

Answer 1

The Correct Option is C

Step 1: Understanding Work Done:
The work done in moving a charge $q$ from point A to point B in the field of charge $Q$ is given by the change in potential energy or $W = q(V_B - V_A)$. However, "work done in taking" usually refers to external work. Usually, Work Done ($W_{ext}$) = $\Delta U = U_B - U_A$. Or, work done *by the field* = $-\Delta U$. Let's calculate the magnitude of energy change first.
Step 2: Formula:
Potential at distance $r$: $V = \frac{kQ}{r}$ \[ W = q(V_B - V_A) = kqQ \left( \frac{1}{r_B} - \frac{1}{r_A} \right) \]
Step 3: Calculation:
Given: $Q = 20 \times 10^{-6}$ C, $q = 2 \times 10^{-6}$ C. $r_A = 0.04$ m, $r_B = 0.05$ m. $k = 9 \times 10^9 \, N m^2 C^{-2}$. \[ W = (9 \times 10^9)(20 \times 10^{-6})(2 \times 10^{-6}) \left( \frac{1}{0.05} - \frac{1}{0.04} \right) \] \[ W = 360 \times 10^{-3} (20 - 25) \] \[ W = 0.36 \times (-5) = -1.8 \, J \] The negative sign indicates that the electric field does positive work (repulsion helps movement from 4cm to 5cm). The external work required to move it *against* the field would be negative (i.e., energy is released or the field pushes it). However, often in such MCQs, the magnitude is asked, or "work done" refers to the energy involved in the process. The magnitude is 1.8 J.
Step 4: Final Answer:
The magnitude of work done is 1.8 J.