The value of one farad in e.s.u. is-

The dimensional formula of capacitance is _______.

Take $Q$ as the dimension formula of charge.

A parallel plate capacitor with plate area '$A$' and distance of separation '$d$' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as:

$\epsilon \left(x\right)={\epsilon }_0+kx$, for $\left(0<x\le \frac{d}{2}\right)$

$\epsilon \left(x\right)={\epsilon }_0+k\left(d-x\right)$, for $\left(\frac{d}{2}\le x\le d\right)$

In the given circuit diagram,

$E=5 \mathrm{V}, r=1 \mathrm{\Omega }, R_2=4 \mathrm{\Omega }, R_1=R_3=1 \mathrm{\Omega }$ and $C=3 \mu \mathrm{F}$

Then, what will be the numerical value of charge on each plates of the capacitor.

An electron enters a parallel plate capacitor with horizontal speed $\mu$ and is found to deflect by angle $\theta$ on leaving the capacitor as shown. It is found that $\tan \theta =0.4$ and gravity is negligible.

If the initial horizontal speed is doubled, then $\tan \theta$ will be:

The material filled between the plates of a parallel plate capacitor has resistivity $200 \Omega \mathrm{m}$. The value of capacitance of the capacitor is $2 \mathrm{pF}$. If a potential difference of $40 \mathrm{V}$ is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permittivity of material is $50$)

Find the total capacitance and total charge on the capacitors

As shown in the figure below, if a capacitor $C$ is charged by connecting it with resistance $R,$ then energy given by the battery will be

In the figure is shown a system of four capacitors connected across a $10 \mathrm{V}$ battery. The charge that will flow from switch S when it is closed is:

What are the charges stored in the $1 \mu F$ and $2 \mu F$ capacitors in the circuit below, once the currents become steady?

Two identical parallel plate capacitors of capacitance each are connected in series with a battery of emf, $E$ as shown. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, the amount of charge which will flow through the battery is (neglect internal resistance of the battery)