Two identical charged spheres are suspended b | vedclass

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(A) Let the angle each string makes with the vertical be $	heta$. Since the total angle between the strings is $30^o$,$	heta = 15^o$.From the free b

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the angle each string makes with the vertical be $	heta$. Since the total angle between the strings is $30^o$,$	heta = 15^o$.From the free body diagram of one sphere in air,the forces are tension $T$,weight $mg$,and electrostatic force $F$. Balancing forces:$T \sin 	heta = F$$T \cos 	heta = mg$Dividing these,we get $	an 	heta = \frac{F}{mg} \quad ......(i)$When suspended in a liquid of density $ho$,the sphere experiences an upward buoyant force $F_B = V ho g$,where $V$ is the volume of the sphere. The effective weight becomes $mg' = mg - F_B = V d g - V ho g = V g (d - ho)$,where $d$ is the density of the sphere.The electrostatic force in the liquid becomes $F' = \frac{F}{K}$,where $K$ is the dielectric constant.Since the angle $	heta$ remains the same:$	an 	heta = \frac{F'}{mg'} = \frac{F/K}{V g (d - ho)} = \frac{F}{K V g d (1 - ho/d)} = \frac{F}{K mg (1 - ho/d)} \quad ......(ii)$Equating $(i)$ and $(ii)$:$\frac{F}{mg} = \frac{F}{K mg (1 - ho/d)}$$1 = \frac{1}{K (1 - ho/d)}$$K = \frac{1}{1 - ho/d} = \frac{1}{1 - (1 / (4/3))} = \frac{1}{1 - 3/4} = \frac{1}{1/4} = 4$.

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