$A$ parallel plate capacitor has circular plates of $0.08\,m$ radius and $1.0 \times 10^{-3}\,m$ separation. If a potential difference of $100\,V$ is applied,the charge on the capacitor will be:

Between the plates of a parallel plate capacitor,there is a $1\,mm$ thick paper of dielectric constant $4$. It is charged to $100\,V$. The electric field in $V/m$ between the plates of the capacitor is:

If the distance between the plates of a parallel plate capacitor of capacity $10 \mu F$ is doubled,then the new capacity will be: (in $\mu F$)

Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant $k$. The charge on each capacitor is $q$. Find the extension in the spring at equilibrium.

The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $y$ from one plate as: $K = \lambda \sec(\pi y/2d)$,where $\lambda$ is a dimensionless constant. The capacitance of this capacitor is

$A$ thin aluminum sheet of negligible thickness is placed between the plates of a parallel plate capacitor. The capacitance of the capacitor: