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```
Q))Two equal  chargs are ixed at x=a and x=-a on d axis.anodr pt charge Q is placed at  d orign.The change in d electric potential energy of Q,when it is  displaced by a small distance x along d x-axis, is approximately  proportional to....
(a)x
(b)x2
(c)x3
(d)1/x
```
7 years ago

Saket Ghanashyam Pande
17 Points
```										Initial p.e.=kqQ/a+kqQ/a=2kqQ/a After small displacement p.e= kqQ/a-x + kqQ/x+a = 2kqQa/a2-x2
Net change=2kqQa/a2-x2 - 2kqQ/a = 2kqQx2/a(a2-x2). Since, x is small x2 can be neglected. So, change=2kqQx2/a(a2)
So,change is proportional to x2.
Note: k=1/4π€
```
7 years ago
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