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 Performed on a Moving Charge in an Electrostatic Field
Provided Information:
Charge at point B (semicircle center): \( q_B = -6 \times 10^{-6} \, C \)
Charge at point D (10 cm from B): \( q_D = +6 \, \mu C \)
Charge in motion: \( q = +5 \times 10^{-6} \, C \)
Semicircle radius: \( r = 0.05 \, \text{m} \)
Trajectory: From point C to point A along the arc
Underlying Principle:
The work done by an electrostatic force is determined solely by the initial and final electric potentials, as electrostatic fields are conservative. The path followed is not a factor.
\[ W = q \Delta V = q (V_A - V_C) \]
1. Electric Potential from a Point Charge:
\[ V = k \frac{q}{r} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)
2. Relevant Distances:
Points A and C are equidistant (5 cm) from B, lying on the semicircle. Thus, the potential due to \( q_B \) is identical at A and C.
Geometrically, the distance from D to A equals the distance from D to C (10 cm). Consequently, the potential due to \( q_D \) is also identical at A and C.
⚡ Critical Observation:
Since the total potentials at A and C are equal ($V_A = V_C$), their difference is zero ($V_A - V_C = 0$).
Zero work is performed when moving the charge \( +5 \, \mu C \) from point C to point A along the semicircle, as the electrostatic potential at both points is the same.
