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(B) Let the four balls be at the corners of a square of side $x$. The distance of each ball from the vertical axis passing through the suspension poin

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Let the four balls be at the corners of a square of side $x$. The distance of each ball from the vertical axis passing through the suspension point is $r = l \sin 45^{\circ} = 20 	imes 10^{-2} 	imes \frac{1}{\sqrt{2}} = \frac{0.2}{\sqrt{2}} \,m$.The distance between two adjacent balls is $x = \sqrt{2} r = 0.2 \,m$.The electrostatic force $F_e$ on one ball due to the other three is the vector sum of forces from the three other charges. Two adjacent charges are at distance $x$,and the diagonal charge is at distance $\sqrt{2} x$.$F_e = \frac{k Q^2}{x^2} \cos 45^{\circ} + \frac{k Q^2}{x^2} \cos 45^{\circ} + \frac{k Q^2}{(\sqrt{2} x)^2} = \frac{2 k Q^2}{x^2} \frac{1}{\sqrt{2}} + \frac{k Q^2}{2 x^2} = \frac{k Q^2}{x^2} (\sqrt{2} + 0.5)$.In equilibrium,$	an 45^{\circ} = \frac{F_e}{mg} \Rightarrow F_e = mg$.$mg = \frac{k Q^2}{x^2} (\sqrt{2} + 0.5)$.Given $m = 0.1 \,kg$,$g = 10 \,m/s^2$,$k = 9 	imes 10^9$,$x = 0.2 \,m$.$0.1 	imes 10 = \frac{9 	imes 10^9 	imes Q^2}{(0.2)^2} (1.414 + 0.5)$.$1 = \frac{9 	imes 10^9 	imes Q^2}{0.04} (1.914)$.$Q^2 = \frac{0.04}{9 	imes 10^9 	imes 1.914} \approx 2.32 	imes 10^{-12} \,C^2$.$Q \approx 1.52 	imes 10^{-6} \,C = 1.52 \,\mu C$.Thus,the value of $Q$ is close to $1.5 \,\mu C$.

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