Two equal point charges are fixed at x = -a a | vedclass

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Question

Potential energy $=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ r }$

$|\Delta P |=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ xdx }-\frac{1}{4

Vedclass · Accepted Answer

$x^2$

Vedclass · Answer

Potential energy $=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ r }$

$|\Delta P |=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ xdx }-\frac{1}{4 \pi \epsilon_0} \frac{ qQ }{ x +d x }$

$\lim _i x \rightarrow 0=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q \cdot 2 x}{x^2-(d x)^2}$

Where $dx =0$

so the answer is $x ^2$

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 $Q$ ehen it is displaced by a small distance $x$ along the $x$ -axis is apporximately proportional to

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