Question

A charge −6 μC is placed at the centre 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 μC is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.

Answer 1

Central charge at B (q_B) = −6 μC = −6 × 10⁻⁶ C

Charge at D (q_D) = +6 μC = +6 × 10⁻⁶ C

Moving charge (q) = +5 μC = +5 × 10⁻⁶ C

Semicircle radius (r) = 5 cm = 0.05 m

The points are arranged horizontally in the order D, C, B, and A.

Distance from B to C (BC) = r = 0.05 m

Distance from B to A (BA) = r = 0.05 m

Distance from D to B (DB) = 10 cm = 0.10 m

Distance from D to C (DC) = DB − BC = 0.10 − 0.05 = 0.05 m

Distance from D to A (DA) = DB + BA = 0.10 + 0.05 = 0.15 m

Both points C and A lie on the circumference.

They are equidistant from the centre B.

V_B,A = V_B,C = `(k * q_B)/r`

Therefore, their potential difference due to q_B is zero.

V_B,A − V_B,C = 0

Using Coulomb’s constant:

k = 9 × 10⁹ N . m²/C²

The potentials at A and C caused by q_D:

V_D,A = `(k * q_D)/(DA)`

= `(9 xx 10^9 xx 6 xx 10^-6)/0.15`

= 3.6 × 10⁵ V

V_D,C = `(k * q_D)/(DC)`

= `(9 xx 10^9 xx 6 xx 10^-6)/0.05`

= 1.08 × 10⁶ V

The net potential difference (ΔV) between the final point A and the initial point C is:

ΔV = V_A − V_C

= (V_B,A − V_B,C) + (V_D,A − V_D,C)

= 0 + 3.6 × 10⁵ − 1.08 × 10⁶

= −7.2 × 10⁵ V

The work done on the moving charge is given by:

W = q . ΔV

= (5 × 10⁻⁶) × (−7.2 × 10⁵)

= −3.6 J