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Two equal point charges are fixed at x=-...

Two equal point charges are fixed at `x=-a` and `x=+a` on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

A

`x`

B

`x^2`

C

`x^3`

D

`1//x`

Text Solution

Verified by Experts

The correct Answer is:
B
Potential enrgy of the system when charge `Q` is at `O` is
`U_0 = (KqQ)/a + (KqQ)/a = (K2 qQ)/a`
When charge `Q` is shifed to position `O` , the potential nergy will be

`U = (KqQ)/((a +x)) + (KqQ)/((a-x))= (KqQ(2a))/((a^2 - x^2 )) = ( 2 KqQ)/a xx (1 - x^2/a^2)^(-1)`
`~~ (2 KqQ)/a ( 1 + x^2 /a^2 ) ( :. x lt lt a)`
`:. Dela U = U - U_0 = (2 KqQ)/a ( 1- z^2 /a^2) - (2 KqQ)/a = (2 KqQ)/a^3 (x^2)`.
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