Two capacitors of capacitance C each are connected in series. The equivalent capacitance is:

A parallel plate capacitor has circular plates of 0.08m radius and $1.0 imes {10^{ - 3}}m$ separation. If a P.D. of $100\;volt$ is applied, the charge will be

64 identical spheres of charge $q$ and capacitance $C$ each are combined to form a large sphere. The charge and capacitance of the large sphere is

Two identical capacitors are charged to potentials $V_1$ and $V_2$ respectively. If they are then connected in parallel with plates of like polarity together, what is the final potential difference across the combination?

A parallel plate capacitor is filled with two dielectrics of permittivities $\epsilon_1$ and $\epsilon_2$, each occupying half the volume as shown. If the area of each plate is A and the plate separation is d, what is the capacitance?

Assertion (A) : From the relation $C=\frac{q}{V}$. We can say that, if more charge $q$ is given to a conductor, its capacitance should increase.

Reason (R) : Ratio $\frac{q}{V}$ will remain constant for a given conductor.

A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is: $\left(ext{Given }\varepsilon_0=8.834 imes 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

The radius of a sphere of capacity $1\,\,\mu F$ in air is

The potentials of the two plates of capacitor are $+ 10V$ and $- 10\;V.$ The charge on one of the plates is $40\;C.$ The capacitance of the capacitor is

When a capacitor is connected to a battery

A cylindrical capacitor is constructed using two coaxial cylinders of the same length $10 \mathrm{~cm}$ and of radii $2 \mathrm{~mm}$ and $4 \mathrm{~mm}$. Calculate the capacitance.