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

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(B) Let the two equal charges be $q$. The potential energy $U$ of charge $Q$ at a position $x$ is given by the sum of potentials due to the two charge

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$x^2$

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(B) Let the two equal charges be $q$. The potential energy $U$ of charge $Q$ at a position $x$ is given by the sum of potentials due to the two charges: $U(x) = k q Q [\frac{1}{a-x} + \frac{1}{a+x}]$,where $k = \frac{1}{4 \pi \epsilon_0}$.Simplifying the expression: $U(x) = k q Q [\frac{a+x+a-x}{a^2-x^2}] = k q Q [\frac{2a}{a^2-x^2}]$.The potential energy at the origin $(x=0)$ is $U(0) = k q Q [\frac{2a}{a^2}] = \frac{2 k q Q}{a}$.The change in potential energy is $\Delta U = U(x) - U(0) = k q Q [\frac{2a}{a^2-x^2} - \frac{2}{a}] = 2 k q Q a [\frac{1}{a^2-x^2} - \frac{1}{a^2}]$.$\Delta U = 2 k q Q a [\frac{a^2 - (a^2-x^2)}{a^2(a^2-x^2)}] = 2 k q Q a [\frac{x^2}{a^2(a^2-x^2)}]$.For small displacement $x$,$a^2-x^2 \approx a^2$. Thus,$\Delta U \approx 2 k q Q a [\frac{x^2}{a^4}] = \frac{2 k q Q}{a^3} x^2$.Therefore,$\Delta U \propto x^2$.

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

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