Two identical charged spheres are suspended b | vedclass

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(C) Let $	heta$ be the angle each string makes with the vertical. Since the total angle between the strings is $37^{\circ}$,$	heta = 37^{\circ}/2 =

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$2$

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(C) Let $	heta$ be the angle each string makes with the vertical. Since the total angle between the strings is $37^{\circ}$,$	heta = 37^{\circ}/2 = 18.5^{\circ}$.In air,the equilibrium condition is $	an 	heta = \frac{F_e}{mg} = \frac{F_e}{ho_B V g}$,where $ho_B$ is the density of the sphere and $V$ is its volume.In the liquid,the effective weight is $mg' = V(ho_B - ho_L)g$ and the electrostatic force becomes $F_e' = \frac{F_e}{K}$,where $K$ is the dielectric constant.The equilibrium condition in the liquid is $	an 	heta = \frac{F_e'}{mg'} = \frac{F_e}{K V(ho_B - ho_L)g}$.Since $	heta$ remains the same,we equate the two expressions for $	an 	heta$:$\frac{F_e}{ho_B V g} = \frac{F_e}{K V(ho_B - ho_L)g}$$ho_B = K(ho_B - ho_L)$$1.4 = K(1.4 - 0.7)$$1.4 = K(0.7)$$K = \frac{1.4}{0.7} = 2$.

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