As shown in the figure, charges `+q and -q` are placed at the vertices `B` and `C` of an isoscles triangle. The potential at the vertex `A` is

Options

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

Answer

B

Explanation

Potential at `A =` Potential due to `(+q)` charge
`+`Potential due to `(-q)` charge
`= (1)/(4pi epsilon_(0)). (q)/(sqrt(a^(2) + b^(2))) + (1)/(4pi epsilon_(0)) ((-q))/(sqrt(a^(2) + b^(2))) = 0`