**Question:**

**Two charges –q each are separated by distance 2d. A third charge +q is kept at midpoint O. Find potential energy of +q as a function of small distance x from O due to –q charges. Sketch PE versus x and convince yourself that the charge at O is in an unstable equilibrium.**

**Solution:**

In the above figure, +q is the charge that got displaced from O towards (-d,0).

This is written as

$U=q\left(V_{1}+V_{2}\right)=q \frac{1}{4 \pi \epsilon_{0}} \frac{-q}{(d-x)}+\frac{-q}{d+x}$

$U=\frac{1}{2 \pi \epsilon_{0}} \frac{-q^{2} d}{d^{2}-x^{2}}$

At x = 0

$U=\frac{1}{2 \pi \epsilon_{0}} \frac{q^{2}}{d}$

Differentiating the equation with respect to x, we get

When x < 0, dU/dx > 0

And when x > 0, dU/dx < 0

Using this we can define, charge on particle to be F = -dU/dx

F = -dU/dx = 0

When

d2U/dx2 = positive, equilibrium is stable

d2U/dx2 = negative, equilibrium is unstable

d2U/dx2 = 0, equilibrium is neutral

Therefore, when x = 0, d2U/dx2 = (-2dq2/4πε0)(1/d6)(2d2) < 0

Which shows that the system is unstable equilibrium.

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.