Subject Experts Solutions for Chapter: Electrostatics, Exercise 1: Practice Exercise

$125$ water drops of equal radii and equal capacitance $C$ coalesce to form a single drop of capacitance $C^{\text{'}}$. The relation between $C$ and $C^{\text{'}}$ is,

A capacitor of capacitance $10 \mu \mathrm{F}$ has been charged to $100 \mathrm{V}$. It is now connected in parallel to another uncharged capacitor $C$. The common potential becomes $40 \mathrm{V}$. The capacitance of the other capacitor $C$ is

A point charge of $5 \mu \mathrm{C}$ is placed at the center of a sphere of radius $9.2 \mathrm{m}$. If ${\epsilon }_0=8.854\times {10}^{-12} {\mathrm{C}}^2 {\mathrm{N}}^{-1} {\mathrm{m}}^{-2}$, then the electric flux through the surface of the sphere will be

Two electric charges $+q$ and $-q$ are situated at a distance apart from each other. The electric field $E$ and the potential $V$ at the midpoint between the charges will be characterised

Two charges in air experience a coulomb force of $20 \mathrm{N}$. If the space between them is filled with a medium of dielectric constant $K=4$, then what is the new force?

An electron of mass $9\times {10}^{-31} \mathrm{kg}$ and charge $1.6\times {10}^{-19} \mathrm{C}$ is placed in an electric field of strength $9\times {10}^6 \mathrm{N} {\mathrm{C}}^{-1}$, the force on electron will be

A capacitor of $20 \mathrm{\mu }\mathrm{F}$ and charged to $500 \mathrm{V}$ is connected in parallel to another capacitor of $10 \mathrm{\mu }\mathrm{F}$ charged to $200 \mathrm{V}$. The common potential will be

Two similarly charged bodies are kept $5 \mathrm{cm}$ apart in air. If the second body is shifted away from the first by another $5 \mathrm{cm}$, their force of repulsion will be