Two balls of same mass and carrying equal cha | vedclass

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(D) In equilibrium,the forces acting on one ball are the electrostatic force $F_e$,tension $T$,and weight $mg$.$F_e = T \sin 	heta$$mg = T \cos 	het

Vedclass · Accepted Answer

$l^{1/3}$

Vedclass · Answer

(D) In equilibrium,the forces acting on one ball are the electrostatic force $F_e$,tension $T$,and weight $mg$.$F_e = T \sin 	heta$$mg = T \cos 	heta$Dividing the two equations,we get $	an 	heta = \frac{F_e}{mg} = \frac{q^2}{4 \pi \epsilon_0 x^2 mg}$.Since the angle $	heta$ is small,$	an 	heta \approx \sin 	heta = \frac{x/2}{l}$.Equating the two expressions for $	an 	heta$:$\frac{x}{2l} = \frac{q^2}{4 \pi \epsilon_0 x^2 mg}$$x^3 = \frac{2 q^2 l}{4 \pi \epsilon_0 mg}$$x^3 = \frac{q^2 l}{2 \pi \epsilon_0 mg}$Thus,$x = \left( \frac{q^2 l}{2 \pi \epsilon_0 mg} ight)^{1/3}$.Therefore,$x \propto l^{1/3}$.

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

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