Capacitors and Capacitance

Calculate final charge on $1\mu \mathrm{F}$ capacitor:

In a capacitor which is initially charged such that the left plate of capacitor has charge $+Q$ and the right plate has zero charge at time $t=0 \mathrm{s}$. The total charge on the right plate after closing the switch will be

Find the potential drop across the capacitor when the circuit given below achieves steady-state.

Calculate the charge on a capacitor of capacitance $C=\left(2.0\pm 0.1\right) \mathrm{\mu F}$ and charged to a voltage $V=\left(20\pm 0.2\right) \mathrm{V}$.

When the charge on a capacitor increases keeping geometry the same

A parallel plate condenser is filled with two dielectrics as shown in the figure. The area of each plate is $A{ \mathrm{m}}^2$ and the separation is d metre. The dielectric constants are $K_1$and $K_2$ respectively. Its capacitance in farad will be-

Effective capacitance of parallel combination of two capacitor $C_1$ and $C_2$ is $10 \mathrm{\mu F}$. When we inserted a dielectric slab between the plates the new capacitance will be ( dielectric constant $K_1=2, K_2=4$)

A parallel plate capacitor of capacitance $15 \mu \mathrm{F}$. A dielectric slab $\left(K=2\right)$ of thickness $1 \mathrm{mm}$ is inserted between the plates. Then new capacitance is given by-

A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $X=0$ and the positive plate is at $X=3d.$ The slab is equidistant from the plates. The capacitor is given some charge. As $X$ goes from $0$ to $3d$ then,

$\left[a\right]$ The electric potential increases at first, then decreases and again increases.
$\left[b\right]$ The electric potential increases continuously.
$\left[c\right]$ The direction of the electric field remains the same
$\left[d\right]$ The magnitude of the electric field remains the same

A capacitor is charged using a battery, and the battery is withdrawn later on. Now a dielectric slab is introduced between the capacitor plates then the correct statement is-

A parallel plate air capacitor is charged by connecting its plates to a battery. Without disconnecting the battery, a dielectric is introduced between its plates. As a result-

The value of a capacitor formed by a thin metallic foil is $2 \mu \mathrm{F}.$ The foil is folded with a layer of paper having a thickness of $0.015 \mathrm{mm}.$ The dielectric constant of the paper is $2.5$ and its breadth is $40 \mathrm{mm}.$ The length of the foil used is-

A sheet of aluminium is inserted in the air gap of a parallel plate capacitor without touching any of the two plates of the capacitor. The capacitance of the capacitor is-

A parallel plate capacitor is first charged and then disconnected from the battery and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

The distance between the plates of a parallel plate capacitor is $d.$ A copper plate of the same area but thickness $\frac{d}{2}$ is placed between the plates then the new capacitance will become-

In the following figure, the charge on each condenser in the steady state will be-

A parallel plate capacitor is charged and then isolated. On increasing the plate separation-

A capacitor of $6 \mathrm{\mu }\text{F}$ is charged to such an extent that the potential difference between the plates becomes $50 \mathrm{V}$. The work done in this process will be,

A $4 \mathrm{\mu }\text{F}$ capacitor is charged to $400 \mathrm{V}$ and then its plates are joined through a resistor of resistance $1 \mathrm{k\Omega }$. The heat produced in the resistor is,

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