Embibe Experts Solutions for Chapter: Electrostatics, Exercise 2: Exercise 2

Two point charge of $8\mu C$ and $12\mu C$ are kept in air at a distance of $10 \mathrm{cm}$ from each other. The work required to change the distance between them to $6 \mathrm{cm}$ will be.

Two parallel plate capacitors of capacitances $C$ and $2C$ are connected in parallel and charged to a potential difference $V.$ The battery is then disconnected and the region between the plates of the capacitor $C$ is completely filled with a material of dielectric constant $K\mathit{.}$ The potential difference across the capacitors now becomes -

The electric potential $V$ as a function of distance $x$ (in metre) is given by: $V=\left(5x^2+10x-9\right)V\mathit{.}$ The value of electric field of $x=1 \mathrm{m}$ would be-

An uncharged capacitor with a solid dielectric is connected to a similar air capacitor charged to a potential of $V_0.$ If the common potential after sharing of charges becomes $V\mathit{,}$ then the dielectric constant of the dielectric must be

Find equivalent capacitance between $X$ and $Y$ if each capacitor is $4 \mathrm{\mu F}$.

The maximum charge stored on a metal sphere of radius $15 \mathrm{cm}$ may be $7.5 \mathrm{\mu C}$. The potential energy of the sphere in this case is

On moving a charge of $20 \mathrm{C}$ by $2 \mathrm{cm}, 2 \mathrm{J}$ of work is done, then the potential difference between the points is

${10}^3$ small water drops, each of radius $r$ and each carrying charge $q,$ combines to form one bigger drop. The potential of the bigger drop as compared to that of smaller drop, will be.