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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

`0`

B

`-(3kq^2)/(2a)`

C

`-(5kq^2)/(2a)`

D

`-(7 kq^2)/(2a)`

Text Solution

Verified by Experts

The correct Answer is:
D
Total energy of system
`v = (kq xx - 2q)/(a) xx 2 + (kqq)/(2a)`
`= - (4kq^2)/(a) + (kq^2)/(2) = - (4kq^2)/(a) + (kq^2)/(2a) = (7 kq^2)/(2a)`
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