Two identically charged pith balls are suspen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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:

Similar Questions

The electrostatic force on a small sphere of charge $0.4 \; \mu C$ due to another small sphere of charge $-0.8 \; \mu C$ in air is $0.2 \; N$.
$(a)$ What is the distance between the two spheres?
$(b)$ What is the force on the second sphere due to the first?

Two equally charged metal spheres $A$ and $B$ repel each other with a force of $4 \times 10^{-5} \ N$. Another identical uncharged sphere $C$ is touched to $A$ and then placed at the midpoint of the line joining the spheres $A$ and $B$. The net electric force on the sphere $C$ is

The sum of two point positive charges separated by a distance of $1.5 \ m$ in air is $25 \mu C$. If the electrostatic force between the two charges is $0.6 \ N$,then the difference between the two charges is (in $\mu C$)

Three identical charges are placed on three vertices of a square. If the force acting between $q_1$ and $q_2$ is $F_{12}$ and between $q_1$ and $q_3$ is $F_{13}$,then $\frac{F_{13}}{F_{12}} = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module