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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

`U_(i)=((2Qq)/(4piepsilon_(0)(a))),U_(f)=(Qq)/(4piepsilon_(0))[(1)/(a+x)+(1)/(a-x)]`
`U_(i)-U_(f)=((Qqx^(2)))/(2piepsilon_(0)^(3))` [for `xltlta,x^(2)` can be neglected in comparison to `a^(2)` in denominator]
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