As per this diagram a point charge `+q` is placed at the origin `O`. Work done in taking another point charge `-Q` from the point `A(0, a)` to another point `B(a, 0)` along the staight path `AB` is:

Options

A. zero
B. `((-qQ)/(4pi epsilon_(0))(1)/(a^(2))) sqrt(2) a`
C. `((-qQ)/(4pi epsilon_(0))(1)/(a^(2))) (q)/(sqrt(2))`
D. `((-qQ)/(4pi epsilon_(0))(1)/(a^(2))) sqrt(2)`

Answer

A

Explanation

Potential at `V_(A) = (1)/(4pi epsilon_(0))(q)/(a)`
Potential at `B V_(B) = (1)/(4pi epsilon_(0))(q)/(a)`
Thus, work done in carrying a test charge `-Q` from `A` to `B`
`W = (V_(A) -V_(B)) (-Q) = 0`