A charge `+Q` at `A` (see figure) produces electric field `E` and electric potential `V` at `D`. If we now put charges `-2Q` and `+Q` at `B` and `C` respectively, then the electric field and potential at `D` will be

Options

A. `E` and `0`
B. `0` and `V`
C. `sqrt(2)E` and `(V)/(sqrt(2))`
D. `(E)/(sqrt(2))` and `(V)/(sqrt(2))`

Answer

A

Explanation

When charges `-2Q` and `+Q` are placed at `B` and `C` respectively,
`V' = (1)/(4pi epsilon_(0)) [(Q)/(a) - (2Q)/(a) + (Q)/(a)] =` zero
This settles in favour of (a).