There are four point charges `+q, -q, +q` and `-q` are placed at the corners `A, B, C` and `D` respectively of a square of side `a`. The potential energy of the system is `(1)/(4pi epsilon_(0))` times

Options

A. `(q^(2))/(a) (-4 + sqrt(2))`
B. `(q^(2))/(2a) (-4 + sqrt(2))`
C. `(4q^(2))/(a)`
D. `(-4sqrt(2)q^(2))/(a)`

Answer

A

Explanation

There will be six pairs,
So, potential energy of the system is
`U = ((q)(-q))/(4pi epsilon_(0)a) + ((-q)(+q))/(4pi epsilon_(0)a) +((+q)(-q))/(4pi epsilon_(0)a)`
`+((-q)(q))/(4pi epsilon_(0)a) +((q)(q))/(4pi epsilon_(0)a sqrt(2)) + ((-q)(-q))/(4pi epsilon_(0)a sqrt(2))`
`U = 4((-q^(2))/(4pi epsilon_(0)a)) + 2(q^(2))/(4pi epsilon_(0)sqrt(2)a)`
`U = (1)/(4pi epsilon_(0)) (q^(2))/(a) (-4 + sqrt(2))`