A charge \( -6 \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \mu C \) is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.
Show Hint
Work done in moving a charge along an equipotential surface is always zero because there is no change in electric potential.
Work Done on a Charge Moving in an Electrostatic Field
Given:
Charge at point B (center of semicircle): \( q_B = -6 \, \mu C = -6 \times 10^{-6} \, C \)
Charge at point D (10 cm from B): \( q_D = +6 \, \mu C \)
Moving charge: \( q = +5 \, \mu C = 5 \times 10^{-6} \, C \)
Radius of the semicircle: \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)
Path: From point C to point A **along the circumference**
Concept:
Work done by the electrostatic force depends only on the **initial and final potential** (electrostatic field is conservative). The path taken is irrelevant.
\[ W = q \left( V_A - V_C \right) \]
1. Potential due to point charge:
\[ V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r} \] For simplicity, use: \[ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \]
2. Distances:
Both points A and C lie on the semicircle: Same distance (5 cm) from B
So, potential due to \( q_B \) is same at A and C
From geometry: Distance from D to A = distance from D to C = 10 cm
So, potential due to \( q_D \) is also same at A and C
⚡ Key Observation:
Since the net potential \( V_A = V_C \), the difference \( V_A - V_C = 0 \)
No work is done in moving the charge \( +5 \, \mu C \) from point C to point A along the semicircle. The electrostatic potential is the same at both points.
