Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula and, Ketki Deshpande Solutions for Chapter: Electrostatics, Exercise 2: Critical Thinking

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux from the top and bottom faces of the cylinder is

The electrostatic energy stored in $1 \mathrm{L}$  volume of air when it is placed in uniform electric field of intensity ${10}^3$ $\mathrm{V} {\mathrm{m}}^{-1}$ is (Use: ${\epsilon }_0=8.85\times {10}^{-12} \mathrm{F} {\mathrm{m}}^{-1}$)

Two parallel metal plates $2.0$ $\mathrm{cm}$ apart, are connected to a $200$ $\mathrm{V}$ battery. A proton with a positive charge $1.6 \times 10^{-19} \mathrm{C}$ is located between the plates. The electric field intensity between the plates is

A capacitor of capacity $4 \mathrm{\mu F}$ charged to $50 \mathrm{V}$ is connected parallel to another capacitor of $2 \mathrm{\mu F}$ charged to $100 \mathrm{V}$ with plates of like charges connected together. What is the total energy before and after connection respectively?

A parallel plate capacitor is connected to a battery. The plates are pulled apart with uniform speed $v$. If $x$ is separation between plates, then the rate of change of the electrostatic energy of the condenser is proportional to

A parallel plate capacitor with air as the dielectric has capacitance $C$. A slab of dielectric constant $k$ and having the same thickness as the separation between the plates is introduced to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

What will be the capacity of a parallel-plate capacitor when the half-of parallel space between the plates is filled by a material of dielectric constant $\varepsilon_{\mathrm{r}} ?$ Assume that the capacity of the capacitor in air is $C$.

Assertion: A parallel plate capacitor is charged by connecting it across a battery. The battery is then disconnected and separation between the plates is decreased as a result of which the energy stored in the capacitor decreases.

Reason: Energy stored in the capacitor is equal to the work done in charging the capacitor.