Two identical non-conducting solid spheres of | vedclass

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

Question

(A) In air,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.At equi

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(A) In air,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.At equilibrium:$T \cos(\alpha/2) = mg$$T \sin(\alpha/2) = F$Dividing the two equations: $	an(\alpha/2) = \frac{F}{mg} = \frac{q^2}{4\pi\epsilon_0 r^2 mg}$.When immersed in a dielectric liquid of dielectric constant $K$,the electrostatic force becomes $F' = \frac{F}{K}$. The sphere also experiences an upward buoyant force $F_B = V ho_l g$,where $V$ is the volume of the sphere and $ho_l$ is the density of the liquid. The effective weight becomes $mg' = mg - V ho_l g = V ho_s g - V ho_l g = V g (ho_s - ho_l)$,where $ho_s$ is the density of the sphere.At equilibrium in the liquid:$T' \cos(\alpha/2) = V g (ho_s - ho_l)$$T' \sin(\alpha/2) = F/K$Dividing the two equations: $	an(\alpha/2) = \frac{F/K}{V g (ho_s - ho_l)}$.Since the angle $\alpha$ remains the same,$	an(\alpha/2)$ is constant:$\frac{F}{mg} = \frac{F/K}{V g (ho_s - ho_l)}$$\frac{1}{mg} = \frac{1}{K V g (ho_s - ho_l)}$Since $m = V ho_s$,we have:$\frac{1}{V ho_s g} = \frac{1}{K V g (ho_s - ho_l)}$$ho_s = K (ho_s - ho_l)$$21 (ho_s - 800) = ho_s$$21 ho_s - 16800 = ho_s$$20 ho_s = 16800 \implies ho_s = 840 \ kg \ m^{-3}$.Since $F' = F/K$,the electric force reduces (Option $B$ is correct). The density of the sphere is $840 \ kg \ m^{-3}$ (Option $C$ is correct).

Similar Questions

The force between two point charges $A$ and $B$ is $F$. If $75\%$ of the charge of $A$ is transferred to $B$,then the new force between $A$ and $B$ is:

Four point charges $q_{A}=2\; \mu C, q_{B}=-5\; \mu C, q_{C}=2\; \mu C,$ and $q_{D}=-5\; \mu C$ are located at the corners of a square $ABCD$ of side $10\; cm$. What is the force on a charge of $1\; \mu C$ placed at the centre of the square (in $; N$)?

Two equal charges $q$ are placed at $x = -a$ and $x = a$ on the $x$-axis. $A$ particle of mass $m$ and charge $q_0 = q/2$ is placed at the origin. If the charge $q_0$ is given a small displacement $(y << a)$ along the $y$-axis,the net force acting on the particle is proportional to .......

Explain the superposition principle for static electric forces and write its general equation.

Two charges $2 C$ and $6 C$ are separated by a finite distance. If a charge of $-4 C$ is added to each of them, the initial force of $12 \times 10^3 \,N$ will change to

Vedclass Test Series

Exam Paper Generator

Online Exam Module