Four charges `+q, +q, -q, and -q` are placed, respectively, at the corners `A,B,C, and D` of a square of side `a`, arranged in the given order. `E and F` are the midpoints of sides `BC and CD`, respectively, `O` is the center of square.
The work done in carrying a charge `e` from `O` to E is.

Options

A. `(sqrt(2) q e)/(pi epsilon_0 a)`
B. ` (q e)/(pi epsilon_0 a)[(1)/(sqrt(5)) - 1]`
C. ` (q e)/(pi epsilon_0 a)[(1)/(sqrt(5)) + 1]`
D. zero

Answer

D

Explanation

Electric field at `O` (Fig. 3.S3.49)
.
`E_0 = 2 (2| vec E_A | cos 45^(@))`
=`4 (1)/(4 pi epsilon_0) q/((a//sqrt2)^2) xx (1)/(sqrt(2))`
=`(sqrt(2) q)/(pi epsilon_0 a^2)`
Electric potential at `O` will be zero. Also `V_E = 0`, so work done in carrying a charge from O to E will be zero.
`V_F = 2 [(1)/(4 pi epsilon_0) q/((a^2 + a^2 // 4)^(1//2)) -(1)/(4 pi epsilon_0) q/(a// 2)]`
=`q/(pi epsilon_0 a) [(1)/(sqrt(5)) - 1]`
Now `W_(O F) = e (V_F - V_O)`.