Two charges of equal magnitude q are placed in air at a distance 2a apart and third charge `-2q` is placed at mid-point . The potential energy of the system is `(epsi_(0)` = permittivity of free space)

Options

A. `-q^(2)/(8piin_(0)a)`
B. `-(3q^(2))/(8piin_(0)a)`
C. `-(5q^(2))/(8piin_(0)a)`
D. `-(7q^(2))/(8piin_(0)a)`

Answer

D

Explanation

We know that work = potential`xx`charge
The work `W_(1)` required to deposit charge q against no field is given by,
`W_(1) = 0`
The work `W_(2)` required to deposit charge q at a distance 2a from the first charge is given by,
`W_(2)=1/(4piepsi_(0))q/(2a)xxq=q^(2)/(4piepsi_(0)xx2a)`
The work `W_(3)` required to deposit charge -2q at a distance a from each charge is given by,
`W_(3)=(V'_(1)+V'_(2))(-2q)`
where `V'_(1)` is potential due to first charge `q_(1) and V'_(2)` is potential due to second charge `q_(2)` at a distance a from each other.
`therefore" "W_(3)=(1/(4piepsi_(0))q/a+1/(4piepsi_(0))q/a)xx-2q`
`=1/(4piepsi_(0))(2q)/axx(-2q)`
`=-(4q^(2))/(4piepsi_(0)a)`
The total work is given by,
`" "W=W_(1)+W_(2)+W_(3)`
`=0+q^(2)/(4piepsi_(0)2a)-(4q^(2))/(4piepsi_(0)a)`
`=-(7q^(2))/(8piepsi_(0)a)`