A capacitor of capacity $0.1 \mu \mathrm{F}$ connected in series to a resistor of $10 \mathrm{~M} \, \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is $ext{(Given, }\log _{10} 2={0 . 3 0 1 0})$

$n$ identical droplets are charged to $V$ volt each. If they coalesce to form a single drop, then its potential will be

When a lamp is connected in series with capacitor, then

A body is charged to a certain potential. When an additional charge of 60 mc is imported to it, the rise in potential is found to be 15 mV. Find the capacitance of the body

Consider a capacitor - charging circuit. Let $Q_1$ be the charge given to the capacitor in time interval of $20 \mathrm{~ms}$ and $Q_2$ be the charge given in the next time interval of $20 \mathrm{~ms}$. Let $10 \mu \mathrm{Ccharge}$ be deposited in a time interval $t_1$ and the next $10 \mu \mathrm{C~charge}$ is deposited in the next time interval $t_2$. Then,

A capacitor of capacity ${C_1}$ is charged upto $V$ volt and then connected to an uncharged capacitor of capacity ${C_2}$. Then final potential difference across each will be

The capacitance of a metallic sphere will be $1\,\mu F,$ if its radius is nearly

When a capacitor is connected to a battery

A capacitor of unknown capacitance $\mathrm{C}$ is connected across a battery of $ext{V}$ volt. The charge stored in it becomes $ext{Q}$ coulomb. When potential across the capacitor is reduced by $ext{V}^{\prime}$ volt, the charge stored in it becomes $ext{Q}^{\prime}$ coulomb. The capacitance $ext{C}$ is

A $500\mu F$ capacitor is charged at a steady rate of $100\mu C/{\rm{second}}.$ The potential difference across the capacitor will be $10\;V$ after an interval of

Assertion:- Increasing the charge on the plates of a capacitor means increasing the capacitance. Reason:- Because $\mathrm{Q}=\mathrm{CV} \Rightarrow \mathrm{Q} \propto \mathrm{C}$.