Two small spheres of each charge q,mass m and | vedclass

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(A) Let $l$ be the length of the thread. In equilibrium,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F_e$. Fr

Vedclass · Accepted Answer

$6 \sqrt{3}$

Vedclass · Answer

(A) Let $l$ be the length of the thread. In equilibrium,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F_e$. From the geometry,$	an 	heta = \frac{F_e}{mg}$,where $	heta$ is the angle the thread makes with the vertical.In air,the angle between the threads is $90^{\circ}$,so $	heta = 45^{\circ}$. The distance between the spheres is $r = 2l \sin 45^{\circ} = l\sqrt{2}$.Thus,$	an 45^{\circ} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2 mg} \Rightarrow 1 = \frac{q^2}{4\pi\epsilon_0 (2l^2) mg} \Rightarrow \frac{q^2}{4\pi\epsilon_0 l^2} = 2mg$ (Eq. $1$).In the liquid,the angle between the threads is $60^{\circ}$,so $	heta' = 30^{\circ}$. The effective weight is $m'g = V(d - ho_{liquid})g = V(d - \frac{2}{3}d)g = \frac{mg}{3}$. The electrostatic force is $F_e' = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{q^2}{r'^2}$,where $r' = 2l \sin 30^{\circ} = l$.Thus,$	an 30^{\circ} = \frac{F_e'}{m'g} \Rightarrow \frac{1}{\sqrt{3}} = \frac{q^2}{4\pi\epsilon_0 \epsilon_r l^2 (mg/3)} \Rightarrow \frac{q^2}{4\pi\epsilon_0 l^2} = \frac{mg}{3\sqrt{3}} \epsilon_r$ (Eq. $2$).Equating Eq. $1$ and Eq. $2$: $2mg = \frac{mg}{3\sqrt{3}} \epsilon_r \Rightarrow \epsilon_r = 6\sqrt{3}$.

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