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Q.

Two balls of same mass and carrying equal charge are hung by threads of length $l$ from a fixed support. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to :

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a

$l^{2}$

b

$l^{2 / 3}$

c

$l$

d

$l^{1 / 3}$

answer is D.

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Detailed Solution

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In equilibrium, $F_{e} = T \sin θ$
$m g = T \cos θ$
$\tan θ = \frac{F_{e}}{m g} = \frac{q^{2}}{4 π \in_{0} x^{2} \times m g}$
Also $\tan θ \approx \sin = \frac{x / 2}{l}$
Hence, $\frac{x}{2 l} = \frac{q^{2}}{4 π \in_{0} x^{2} \times m g}$
$\Rightarrow x^{3} = \frac{2 q^{2} l}{4 π \in_{0} m g}$
$∴ x = {(\frac{q^{2} l}{2 π \in_{0} m g})}^{1 / 3}$
Therefore $x α l^{1 / 3}$

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