A charge `+q` is placed at the origin O of X- Y axes as shown in the figure . The work done in taking a charge Q from A to B along the straight line AB is

Options

A. `(qQ)/(4pi epsilon_0)((a-b)/(ab))`
B. `(qQ)/(4pi epsilon_0)((b-a)/(ab))`
C. `(qQ)/(4pi epsilon_0)(b/a^2-1/b)`
D. `(qQ)/(4pi epsilon_0)(a/b^2-1/b)`

Answer

A

Explanation

`W=QV_(BA)=Q(V_B-V_A)`
`=Q[(Kq)/b-(Kq)/a]=(Qq)/(4piepsilon_0)[(a-b)/(ab)]`