Two identical charged spheres are suspended b | vedclass

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

Question

(D) Let $ho$ be the density of the sphere and $\sigma$ be the density of the liquid. When the spheres are in air,the equilibrium condition is $	an

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Let $ho$ be the density of the sphere and $\sigma$ be the density of the liquid. When the spheres are in air,the equilibrium condition is $	an 	heta = \frac{F_e}{mg}$.
When immersed in a liquid,the effective weight becomes $mg' = V(ho - \sigma)g$ 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'}{m'g} = \frac{F_e / K}{V(ho - \sigma)g}$.
Since the angle $	heta$ remains the same,we equate the two expressions for $	an 	heta$:
$\frac{F_e}{Vho g} = \frac{F_e}{K V(ho - \sigma)g}$.
This simplifies to $K = \frac{ho}{ho - \sigma}$.
Given $ho = 1.6 \, g \cdot cm^{-3}$ and $\sigma = 0.8 \, g \cdot cm^{-3}$,we get $K = \frac{1.6}{1.6 - 0.8} = \frac{1.6}{0.8} = 2$.

Similar Questions

Identify the wrong statement in the following. Coulomb's law correctly describes the electric force that

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is

$ABC$ is a right-angled triangle in which $AB = 3 \ cm$,$BC = 4 \ cm$ and the right angle is at $B$. Three charges $+15 \ \mu C$,$+12 \ \mu C$ and $-20 \ \mu C$ are placed respectively at $A$,$B$ and $C$. The force acting on the charge at $B$ is (in $N$)

$A$ point charge $Q$ is placed at the center of the line joining two equal point charges $+q$ and $+q$. The value of $Q$ if the system of the charges is in equilibrium,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module