First slide

Potential energy

Question

Three point charges of equal value q are placed at the vertices of al equilateral triangle. The resulting lines of force should be sketch as in

Easy

A
B
C
D

Solution

When Q is at position x, then
$\begin{array}{l} U = \frac{qQ}{4 {πε}_{0}} [\frac{1}{(a + x)} + \frac{1}{(a - x)}] \\ = \frac{2 aqQ}{4 {πε}_{0} (a^{2} - x^{2})} \end{array}$
$∴ \frac{dU}{dx} = \frac{2 aqQ}{4 {πε}_{0}} [\frac{2 x}{{(a^{2} - x^{2})}^{2}}] = \frac{2 aqQ}{4 {πε}_{0}} \times \frac{2 x}{a^{4}}$
$(∵ a^{2} - x^{2} = a^{2} when x is small)$
or $dU \propto xdx or U \propto x^{2}$ .

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