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(A) In air,the forces acting on the pith ball are tension $T$,weight $mg$,and electrostatic force $F_0$. Resolving the forces:$T \sin 	heta = F_0$ $(

Vedclass · Accepted Answer

$\frac{\rho}{\rho - \sigma}$

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(A) In air,the forces acting on the pith ball are tension $T$,weight $mg$,and electrostatic force $F_0$. Resolving the forces:$T \sin 	heta = F_0$ $(i)$$T \cos 	heta = mg$ (ii)Dividing $(i)$ by (ii),we get $	an 	heta = \frac{F_0}{mg}$,so $F_0 = mg 	an 	heta$.When immersed in a medium of dielectric constant $K$ and density $\sigma$,the effective weight of the ball becomes $mg' = mg - F_B = mg - V\sigma g = Vho g - V\sigma g = Vg(ho - \sigma) = mg(1 - \frac{\sigma}{ho})$,where $V$ is the volume of the ball.The new electrostatic force is $F_m = \frac{F_0}{K}$.Since the angle $	heta$ remains the same:$	an 	heta = \frac{F_m}{mg'} = \frac{F_0 / K}{mg(1 - \frac{\sigma}{ho})}$Equating the two expressions for $	an 	heta$:$\frac{F_0}{mg} = \frac{F_0}{K mg (1 - \frac{\sigma}{ho})}$$1 = \frac{1}{K (1 - \frac{\sigma}{ho})}$$K = \frac{1}{1 - \frac{\sigma}{ho}} = \frac{ho}{ho - \sigma}$

Two identically charged pith balls are suspended from the same point by two massless identical threads. The density of each ball is $\rho$. If the system is immersed in a medium of density $\sigma$ and the balls remain at the same angle of deflection,then the dielectric constant of the medium is:

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