First slide

Electrostatic force

Question

In the given figure two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal length L. Assume that $θ$ is so small that $t a n θ \approx s i n θ$ then for equilibrium x is equal to

Moderate

A

${(\frac{q^{2} L}{2 π ε_{0} m g})}^{\frac{1}{3}}$
B

${(\frac{q L^{2}}{2 π ε_{0} m g})}^{\frac{1}{3}}$
C

${(\frac{q^{2} L^{2}}{4 π ε_{0} m g})}^{\frac{1}{3}}$
D

${(\frac{q^{2} L}{4 π ε_{0} m g})}^{\frac{1}{3}}$

Solution

In equilibrium
$\begin{array}{c} F_{e} = T \sin θ . . . . . . . . . . . . . . (i) \\ m g = T \cos θ . . . . . . . . . . . . . (i i) \end{array}$
$\begin{array}{l} \tan θ = \frac{F_{e}}{m g} = \frac{q^{2}}{4 π ε_{o} x^{2} \times m g} \\ also \tan θ \approx \sin θ = \frac{x / 2}{L} \\ Hence \frac{x}{2 L} = \frac{q^{2}}{4 π ε_{o} x^{2} \times m g} \end{array}$
$\Rightarrow x^{3} = \frac{2 q^{2} L}{4 π ε_{o} m g} \Rightarrow x = {(\frac{q^{2} L}{2 π ε_{o} m g})}^{1 / 3}$

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