(A) Let the capacitance of the capacitor be $C$ and the $EMF$ of the battery be $V$.When the capacitor is fully charged,the charge stored on the plate

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(A) Let the capacitance of the capacitor be $C$ and the $EMF$ of the battery be $V$.When the capacitor is fully charged,the charge stored on the plates is $Q = CV$.The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.The work done by the battery in charging the capacitor is $W = Q \times V = (CV) \times V = CV^2$.The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} CV^2}{CV^2} = \frac{1}{2} = 0.5$.

Class 12 Physics Electric Potential and Capacitance Energy Stored in a Capacitor

Medium

When a parallel plate capacitor is fully charged by a battery,the potential difference between the two plates becomes equal to the electromotive force $(EMF)$ of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

A
$0.5$
B
$1$
C
$2$
D
$0.25$

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Electric Potential and Capacitance

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Energy Stored in a Capacitor

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When a parallel plate capacitor is fully charged by a battery,the potential difference between the two plates becomes equal to the electromotive force $(EMF)$ of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

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Work done in placing a charge of $8 \times 10^{-18} \text{ C}$ on a capacitor of capacitance $800 \mu\text{F}$ is . . . . . . .

If $C$ is the capacitance and $V$ is the electric potential,then the dimensional formula of $C V^{2}$ is

The heat generated through $2 \, \Omega$ and $8 \, \Omega$ resistances separately, when a capacitor of $200 \, \mu F$ capacity charged to $200 \, V$ is discharged one by one, will be

$A$ parallel plate capacitor has a uniform electric field $\overrightarrow{E}$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A$,the energy stored in the capacitor is: ($\varepsilon_{0} =$ permittivity of free space)

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