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Question

(C) The work done by the electric field in moving a charge $q_0$ from point $A$ to point $B$ is given by $W_{AB} = q_0(V_A - V_B)$.Similarly,the work

Vedclass · Accepted Answer

$W_{AB} = W_{AC} 
eq 0$

Vedclass · Answer

(C) The work done by the electric field in moving a charge $q_0$ from point $A$ to point $B$ is given by $W_{AB} = q_0(V_A - V_B)$.Similarly,the work done in moving the charge from $A$ to $C$ is $W_{AC} = q_0(V_A - V_C)$.Since $B$ and $C$ lie on the circumference of a circle with the charge $q$ at its center,both points are at the same distance $r$ from the charge $q$.Therefore,the electric potential at $B$ and $C$ is the same,i.e.,$V_B = V_C = \frac{kq}{r}$.Since $V_B = V_C$,it follows that $W_{AB} = q_0(V_A - V_B) = q_0(V_A - V_C) = W_{AC}$.Since $A$ is outside the circle,$V_A 
eq V_B$,so $W_{AB} = W_{AC} 
eq 0$.

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