Four identical pendulums are made by attachin | vedclass

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Question

(b)

Electrostatic force on any of the ball is (let $x=$ separation between two adjacent balls).

$F_e=\frac{k q^2}{(\sqrt{2} x)^2}+\frac{2 k q}{x

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(b)

Electrostatic force on any of the ball is (let $x=$ separation between two adjacent balls).

$F_e=\frac{k q^2}{(\sqrt{2} x)^2}+\frac{2 k q}{x^2} \cdot \cos 45^{\circ}$

$=\frac{k q^2}{2 x^2}+\sqrt{2}=\frac{k q}{x^2}$

$=\left(\frac{1}{2}+\sqrt{2}ight) \frac{k q^2}{x^2} \approx \frac{2 k q^2}{x^2}$

As each ball is at an angle of $45^{\circ}$ from each other, so in equilibrium, we have

$	an 45^{\circ} =\frac{F_e}{m g}$

$\Rightarrow m g =F_e$

$\Rightarrow m g =\frac{2 k q^2}{x^2}$

$\Rightarrow m g x^2 =2 k q^2$

$	ext { where, } x=l \sin 	heta =\frac{l}{\sqrt{2}}$

So, substituting values, we get

$\Rightarrow 100 	imes 10^{-3} 	imes 10 	imes\left(\frac{20 	imes 10^{-2}}{\sqrt{2}}ight)^2$ $=2 	imes 9 	imes 10^9 	imes q^2$

$\Rightarrow q^2=\frac{10^{-2}}{9 	imes 10^9}$

$\Rightarrow q^2 > 10^{-12}$ (slightly higher than $10^{-12}$ )

$\Rightarrow 10^{-6} \,C$

So, nearest answer is $1.5\,\mu C .$

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