Embibe Experts Solutions for Chapter: Capacitance, Exercise 1: Exercise#1

A capacitor of capacitance $0.1 \mathrm{\mu F}$ is connected to a battery of $\text{emf}$ $8 \mathrm{V}$ as shown in the figure. Under steady state condition.

Metallic sphere of radius $R$ is charged to potential $V$. Then charge $q$ is proportional to, 

The capacitance of the capacitors of plate areas $A_1$ and $A_2\left(A_1<A_2\right)$ at a distance $d$ is,

Two capacitors $3 \mathrm{\mu }\text{F}$ and $4 \mathrm{\mu }\text{F}$ are individually charged across a $6 \mathrm{V}$ battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the common potential?

Two spherical conductors $A$ and $B$ of radii $a$ and $b \left(b>a\right)$ are placed concentrically in air. $A$ is given a charge $+Q$ while $B$ is earthed. Then the equivalent capacitance of the system is,

A parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result, the potential difference between the plates, 

Force acting upon a charged particle kept between the plates of a charged condenser is $F$. If one of the plates of the condenser is removed, force acting on the same particle becomes

A capacitor of $6 \mathrm{\mu }\text{F}$ is charged to such an extent that the potential difference between the plates becomes $50 \mathrm{V}$. The work done in this process will be,