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(D) In air,the equilibrium condition is given by $	an(	heta/2) = \frac{F}{mg} = \frac{q^2}{4 \pi \varepsilon_0 r^2 mg}$.When suspended in a liquid o

Vedclass · Accepted Answer

$3$

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(D) In air,the equilibrium condition is given by $	an(	heta/2) = \frac{F}{mg} = \frac{q^2}{4 \pi \varepsilon_0 r^2 mg}$.When suspended in a liquid of dielectric constant $K$ (or $\varepsilon_r$),the electrostatic force becomes $F' = \frac{F}{K}$ and the effective weight becomes $mg' = mg(1 - \frac{ho_{liquid}}{ho_{sphere}})$.Given that the angle $	heta$ remains the same,we have $	an(	heta/2) = \frac{F'}{mg'} = \frac{F/K}{mg(1 - ho_{liquid}/ho_{sphere})}$.Equating the two expressions for $	an(	heta/2)$:$\frac{F}{mg} = \frac{F}{K mg (1 - ho_{liquid}/ho_{sphere})}$$1 = \frac{1}{K (1 - 1/1.5)}$$K = 1 - 1/1.5 = 1 - 2/3 = 1/3$ is incorrect. Let's re-evaluate: $K = \frac{ho_s}{ho_s - ho_l} = \frac{1.5}{1.5 - 1} = \frac{1.5}{0.5} = 3$.Thus,the dielectric constant of water is $3$.

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