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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
`U_(i)=(2KQq)/(a)+(K.q.q)/(2a)`
and `U_(f)-KQq[(1)/(a+x)+(1)/(a-x)]+(K.q.q)/(2a)`
Here, `K=(1)/(4piepsi_(0))," "DeltaU=U_(f)-U_(i)" ""or"" "|DeltaU|=(2KQqx^2)/(a^3)`
For `xltlta" "therefore" "DeltaUpropx^(2)`
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