In the following diagram,the work done in mov | vedclass

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(A) The electric potential $V$ at any point at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.Since points $A, B$ and $C$ lie o

Vedclass · Accepted Answer

$W_A = W_B = W_C 
eq 0$

Vedclass · Answer

(A) The electric potential $V$ at any point at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.Since points $A, B$ and $C$ lie on a semicircle with the charge $q$ at its centre,all these points are at the same distance $r$ (the radius of the semicircle) from the charge $q$.Therefore,the electric potential at points $A, B$ and $C$ is the same,i.e.,$V_A = V_B = V_C = V$.The work done in moving a charge $q_0$ from point $P$ to a point $X$ is given by $W = q_0(V_X - V_P)$.Thus,the work done to move the charge to points $A, B$ and $C$ are:$W_A = q_0(V_A - V_P) = q_0(V - V_P)$$W_B = q_0(V_B - V_P) = q_0(V - V_P)$$W_C = q_0(V_C - V_P) = q_0(V - V_P)$Since $V_A = V_B = V_C$,it follows that $W_A = W_B = W_C$.Since point $P$ is at a different distance from the charge $q$ compared to the points on the semicircle,$V_P 
eq V$,therefore $W_A = W_B = W_C 
eq 0$.

In the following diagram,the work done in moving a point charge from point $P$ to points $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. Then ($A, B, C$ are points on a semicircle and point charge $q$ is at the centre of the semicircle):

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