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  • 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.

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.

 

Answers (1)

Explanation:-

 

In the above figure, +q the charge is at a point away from O towards \left ( -d,0 \right ).

This can be written as

U=q\left ( V_{1}+V_{2} \right )=q\frac{1}{4\pi\; \varepsilon _{0}}[\frac{-q}{\left ( d-x \right )}+\frac{-q}{d+x}]

U=\frac{1}{2\pi\varepsilon _{0}}\frac{-q^{2}d}{d^{2}-x^{2}}

At x=0;

U=\frac{1}{2\pi\varepsilon _{0}}\frac{q^{2}}{d}

If we differentiate both sides of the equation with respect to x, we get

\frac{dU}{dx}>0, when x<0

and \frac{dU}{dx}<0, when x>0

With the help of these two, we can assume that the charge on the particle be

F=\frac{-dU}{dx}

Hence, F=\frac{-dU}{dx}=0

When

a) The condition is given, 

     \frac{d^2U}{dx^{2}} = positive, equilibrium is stable

b) \frac{d^2U}{dx^{2}} = negative, equilibrium is unstable

c) \frac{d^2U}{dx^{2}}=0, equilibrium is neutral

Therefore, when

x=0,\frac{d^2U}{dx^{2}}=\left ( \frac{-2dq2}{4\pi\varepsilon _{0}} \right )\left ( \frac{1}{d^{6}} \right )\left ( 2d^2 \right )<0

Which shows that the system is an unstable equilibrium.

Posted by

infoexpert23

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