Question

Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis. The charge q at origin in equilibrium will have

• A. Maximum force and minimum potential energy
• B. Minimum force & maximum potential energy
• C. Maximum force & maximum potential energy
• D. Minimum force & minimum potential energy.

Answer 1

Answer: (D)

Minimum force & minimum potential energy.

Answer 2

The net force on q at origin is

$\overrightarrow{F} = {\overrightarrow{F}}_{1} + {\overrightarrow{F}}_{2}$ = $\frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{r^{2}}\widehat{i} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{r^{2}}( - \widehat{i}) = \widehat{0}$

The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is

U = $\frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{(a - x)} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{(a + x)}$ =

$\frac{1}{4\pi\varepsilon_{0}}.q^{2}\left\lbrack \frac{1}{a - x} + \frac{1}{a + x} \right\rbrack$.
$\frac{dU}{dx} = \frac{q^{2}}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{(a - x)^{2}} - \frac{1}{(a + x)^{2}} \right\rbrack$

For U to be minimum, $\frac{dU}{dx} = 0,$ $\frac{d^{2}U}{dx^{2}} > 0,$

⇒ (a-x)² = (a + x)² ⇒ a + x = ± (a – x)

⇒ x = 0, because other solution is irrelevant.

Thus the charged particle at the origin will have minimum force and minimum P.E.