A metal ball of radius R is placed concentric | vedclass

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

Question

(A) Let the inner ball be $S_1$ (radius $R$,charge $q_1 = +2Q$).Let the hollow sphere be $S_2$ (inner radius $r_2 = 2R$,outer radius $r_3 = 3R$,total

Vedclass · Accepted Answer

$\frac{7Q^2}{24\pi\varepsilon_0 R}$

Vedclass · Answer

(A) Let the inner ball be $S_1$ (radius $R$,charge $q_1 = +2Q$).Let the hollow sphere be $S_2$ (inner radius $r_2 = 2R$,outer radius $r_3 = 3R$,total charge $q_2 = -Q$).Due to electrostatic induction,a charge of $-2Q$ is induced on the inner surface of the hollow sphere $(r=2R)$.Since the total charge on the hollow sphere is $-Q$,the charge on its outer surface $(r=3R)$ must be $q_{out} = -Q - (-2Q) = +Q$.The electrostatic potential energy $U$ of a system of charged conductors is given by $U = \frac{1}{2} \sum q_i V_i$.The potential of the inner ball $(V_1)$ is due to its own charge and the charges on the hollow sphere:$V_1 = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{R} + \frac{-2Q}{2R} + \frac{Q}{3R} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{R} - \frac{Q}{R} + \frac{Q}{3R} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{R} + \frac{Q}{3R} \right) = \frac{4Q}{12\pi\varepsilon_0 R} = \frac{Q}{3\pi\varepsilon_0 R}$.The potential of the hollow sphere $(V_2)$ is the same at its inner and outer surfaces:$V_2 = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{2R} + \frac{-2Q}{2R} + \frac{Q}{3R} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{R} - \frac{Q}{R} + \frac{Q}{3R} \right) = \frac{Q}{12\pi\varepsilon_0 R}$.Now,$U = \frac{1}{2} (q_1 V_1 + q_2 V_2) = \frac{1}{2} \left( (2Q) \cdot \frac{Q}{3\pi\varepsilon_0 R} + (-Q) \cdot \frac{Q}{12\pi\varepsilon_0 R} \right)$.$U = \frac{1}{2} \left( \frac{2Q^2}{3\pi\varepsilon_0 R} - \frac{Q^2}{12\pi\varepsilon_0 R} \right) = \frac{1}{2} \left( \frac{8Q^2 - Q^2}{12\pi\varepsilon_0 R} \right) = \frac{1}{2} \left( \frac{7Q^2}{12\pi\varepsilon_0 R} \right) = \frac{7Q^2}{24\pi\varepsilon_0 R}$.

$A$ metal ball of radius $R$ is placed concentrically inside a hollow metal sphere of inner radius $2R$ and outer radius $3R$. The ball is given a charge $+2Q$ and the hollow sphere is given a total charge $-Q$. The electrostatic potential energy of this system is:

Similar Questions

$A$ circuit is shown in the figure for which $C_1=(3 \pm 0.011) \mu F$,$C_2=(5 \pm 0.01) \mu F$,and $C_3=(1 \pm 0.01) \mu F$. If $C$ is the equivalent capacitance across $AB$,then $C$ is given by:

In the given circuit, what are the potential differences between points $A$ and $B$ $(V_{AB})$ and between points $B$ and $C$ $(V_{BC})$?

Four capacitors and a battery are connected as shown. The potential drop across the $7 \mu F$ capacitor is $6 \ V$. Then the:

Minimum number of $2\,\mu F$ and $250\, V$ capacitors required to make a combination of $12\,\mu F$ and $500\, V$ is:

$A$ parallel plate capacitor with a dielectric of dielectric constant $K$ between the plates has a capacitance $C$ and is charged to a potential $V \ volt$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Vedclass Test Series

Exam Paper Generator

Online Exam Module