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Two charges of equal magnitude q are pla...

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)

A

`-(q^(2))/(8piepsi_(0)a)`

B

`-(3q^(2))/(8piepsi_(0)a)`

C

`-(5q^(2))/(8piepsi_(0)a)`

D

`-(7q^(2))/(8piepsi_(0)a)`

Text Solution

Verified by Experts

The correct Answer is:
D

Electric potential energyy of the system in unit volume is
`U=(1)/(4piepsi_(0))((q)(-2q))/(a)+(1)/(4piepsi_(0))((q)(q))/(2a)`
`=(1)/(4piepsi_(0))((q)(-2q))/(a)`
`=(1)/(4piepsi_(0))[(-2q^(2))/(a)+(q^(2))/(2a)+(-2q^(2))/(a)]`
`=(1)/(4piepsi_(0))[(-4q^(2)+q^(2)-4q^(2))/(2a)]`
`=-(7q^(2))/(8piepsi_(0)a)`
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