Question

Figure shows an electric line of force which curves along a circular arc. The magnitude of electric field intensity is same at all points on this curve and is equal to E. If the potential at A is V, then the potential at B is-

• A. V – ERq
• B. V – 2ER sin$\frac{\theta}{2}$
• C. V + ERq
• D. V + 2ER sin $\frac{\theta}{2}$

Answer 1

Answer: (A)

V – ERq

Answer 2

V_A – V_B = work done by electric field on + q coulomb.

Charge from A to B = ERq

\ V_B = V_A – ERq = V – ERq