Two equal point charges are fixed at and on the -axis. Another point charge is placed at the origin. The change in the electrical potential energy of , when it is displaced by a small distance along the -axis, is approximately proportional to

Question

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 [2002]

Smart Achievers · Accepted Answer

(2)

Initial energy

$U_{i} = \frac{2 Q q}{4 π ε_{0} a} + \frac{1}{4 π ε_{0}} \frac{q^{2}}{2 a}$

Final energy

$U_{f} = \frac{Q q}{4 π ε_{0}} [\frac{1}{a + x} + \frac{1}{a - x}] + \frac{1}{4 π ε_{0}} \frac{q^{2}}{2 a} = \frac{2 Q q a}{4 π ε_{0} (a^{2} - x^{2})}$

$∴$ $U_{i} - U_{f} = \frac{2 Q q}{4 π ε_{0}} [\frac{1}{a} - \frac{a}{(a^{2} - x^{2})}]$

$= \frac{2 Q q}{4 π ε_{0}} [\frac{a^{2} - x^{2} - a^{2}}{a (a^{2} - x^{2})}] = \frac{- 2 Q q x^{2}}{4 π ε_{0} a^{3}}$

When $x ≪ a$ , then $x^{2}$ is neglected in the denominator.

$U_{i} - U_{f} = (\frac{- Q q}{2 π ε_{0} a^{3}}) x^{2}$

$∴$ $U_{i} - U_{f} \propto x^{2}$

Solution

Q.
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 [2002]

1 $x$

2 $x^{2}$

3 $x^{3}$

4 $1 / x$

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Solution

Q. 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 [2002]

1 x

2 x2

3 x3

4 1/x