Energy of Charged Capacitor

An uncharged capacitor of capacitance $100 \mathrm{\mu F}$ is connected to a battery of emf$20 \mathrm{V}$ at $t=0$ through a resistance of $10 \mathrm{\Omega }$, then:

(i) The maximum rate at which energy is stored in the capacitor

(a) $10 \mathrm{J} {\mathrm{s}}^{-1}$

(b)  $20 \mathrm{J} {\mathrm{s}}^{-1}$

(c) $40 \mathrm{J} {\mathrm{s}}^{-1}$

(d) $5 \mathrm{J} {\mathrm{s}}^{-1}$

(ii) Time at which the rate has this maximum value

(a) $\left(4\ln 2\right) \mathrm{ms}$

(b) $\left(2\ln 2\right) \mathrm{ms}$

(c) $\left(\ln 2\right) \mathrm{ms}$

(d) $\left(3\ln 2\right) \mathrm{ms}$

A $600 \mathrm{pF}$ capacitor is charged by $200 \mathrm{V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 \mathrm{pF}$ capacitor. Electrostatic energy lost in the process is _____ $\mathrm{\mu J}$.

In the circuit shown, the energy stored in the capacitor is $n \mathrm{\mu J}$. The value of $n$ is _____.

A parallel plate capacitor of capacitance $2 \mathrm{F}$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $\frac{E_2}{E_1}$ is

Find the energy stored in the capacitor in the given circuit.

Condenser is a device used to store

A parallel plate capacitor is charged by connecting a $2 \mathrm{V}$ battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pairs of quantities decrease?

A parallel plate capacitor of capacitance $8 \mu \mathrm{F}$ is connected to a $20 \mathrm{V}$ battery and is allowed to charge completely. The battery is then disconnected and a dielectric material of dielectric constant $8$ is introduced between the plates of the capacitor. The energy dissipated in this process is

A $5.0 \mu \mathrm{F}$ capacitor is charged by a $2.0 \mathrm{V}$ battery. It is then disconnected from the battery and is connected to another uncharged $5.0 \mu \mathrm{F}$ capacitor. The loss in the electrostatic energy in the process is

A $10 \mu \mathrm{F}$ capacitor is charged by $200 \mathrm{V}$. The heat lost in charging it is

If capacitance is increased by $2$ times what will be the energy? Give the answer for vice versa case also?

Calculate the equivalent capacity of the system as shown in the figure. What is the energy stored in the system if voltage applied is $500 \mathrm{V}$.

The capacitance of system is $C$. If key is closed, the total energy loss is equal to:

The total electrostatic energy stored in both the capacitors (in $\mathrm{\mu J}$) is

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates with voltage with voltage source remaining connected to the capacitor. Then

A parallel plate capacitor is connected to a battery which maintains a potential difference $V$ between its plates. A slab of dielectric constant $3$ is inserted between the plates of the capacitor filling the space between them. If $Q$ was the charge on the capacitor plates before insertion of the dielectric, the work done by the battery during the process of insertion of the dielectric is

A capacitor with capacitance $C$ is connected to a battery with voltage $V$ till it is fully charged. The electrical energy stored in the capacitor is $U$. The capacitor is then disconnected from the battery and connected in parallel to an uncharged capacitor with capacitance $2C$ in a parallel combination. The final electrical energy of the combination is

A commonly used (AAA) rechargeable battery has a voltage of $1.2 \mathrm{V}$ and is of$1000 \mathrm{mAh}$ (milli Ampere Hours) capacity. A capacitor has a capacitance of $100 \mathrm{F}$ and can be charged to $2.5 \mathrm{V}$$.$ It follows that

A parallel plate capacitor having plate area $25{ \mathrm{cm}}^2$ and separation $1 \mathrm{mm}$ is connected to a battery of $6 \mathrm{V}$, the work done by the battery during the process is

A capacitor of capacitance $5 \mathrm{\mu F}$ is connected to a source of constant emf of $200 \mathrm{V}$ for a long time, then the switch was shifted to contact $1$ from contact $2$ . The amount of heat generated in the $500 \mathrm{\Omega }$ resistance is $H$. Find value of $16H$ (in $\mathrm{J}$).