K K Sharma Solutions for Chapter: Electrostatic Potential and Capacitance, Exercise 1: TOPICWISE QUESTIONS

A parallel plate condenser is charged to a certain potential and then, disconnected. The separation of the plates is now increased by $2.4 \mathrm{mm}$ and a plate of thickness $3 \mathrm{mm}$ is inserted into it keeping its potential constant. Find the dielectric constant of the medium.

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

A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant, $K_1=3$ and thickness $\frac{d}{3}$ while the other one has a dielectric constant, $K_2=6$ and thickness $\frac{2d}{3}$. Capacitance of the capacitor is now

Due to polarisation of a dielectric slab filling the gap between capacitor plates, the charge induced is $80\%$ that of the charge appearing on the capacitor plate. The dielectric constant of the slab material is

A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is $U$. Now, the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of electric field in pulling out the insulator is $4U.$ Then, the dielectric constant of the insulator is 

A slab of copper of thickness $\frac{d}{2}$ is introduced between the plates of a parallel plate capacitor which touches one of the plates of the capacitor, where $d$ is the separation between its two plates. If the capacitance of the capacitor without copper slab is $C$ and with copper slab is $C^{\text{'}}$, then $\frac{C^{\text{'}}}{C}$ is

The capacity of a parallel plate air capacitor is $10 \mathrm{\mu F}$. As shown in the figure, this capacitor is divided into two equal parts. These parts are filled by media of dielectric constants $K_1=2$ and $K_2=4.$ The capacity of this arrangement will be

The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10 \mathrm{mm}$. There are two dielectric sheets in it, one of dielectric constant $10$ and thickness $6 \mathrm{mm}$ and the other of dielectric constant $5$ and thickness $4 \mathrm{mm}$. The capacity of the condenser is