If $Q$ is the charge on the plates of a capacitor of capacitance $C, V$ the potential difference between the plates, $A$ the area of each plate and $d $ the distance between the plates, the force of attraction between the plates is
$\frac{1}{2}\,\,\left( {\frac{{{Q^2}}}{{{\varepsilon _0}\,A}}} \right)$
$\frac{1}{2}\,\,\left( {\frac{{C{V^2}}}{d}} \right)$
$\frac{1}{2}\,\,\left( {\frac{{C{V^2}}}{{A\,{\varepsilon _0}}}} \right)$
$A$ and $B$ both
A parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, volt$ battery. Then the change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ and the plates are maintained in connection with the battery, is
$27$ similar drops of mercury are maintained at $10 \,V$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ....... times that of a smaller drop.
A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates, the time rate of change of electrostatic energy of capacitor is proportional to
A condenser of capacity ${C_1}$ is charged to a potential ${V_0}$. The electrostatic energy stored in it is ${U_0}$. It is connected to another uncharged condenser of capacity ${C_2}$ in parallel. The energy dissipated in the process is
In a charged capacitor, the energy resides