(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.Given:Capacitance $C = 4 \times 10^{-6} \ F$Potential difference $

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(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.Given:Capacitance $C = 4 \times 10^{-6} \ F$Potential difference $V = 100 \ V$Substituting the values into the formula:$U = \frac{1}{2} \times (4 \times 10^{-6}) \times (100)^2$$U = \frac{1}{2} \times 4 \times 10^{-6} \times 10000$$U = 2 \times 10^{-6} \times 10^4$$U = 2 \times 10^{-2} \ J$$U = 0.02 \ J$Therefore,the energy required to discharge the capacitor is $0.02 \ J$.

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

Easy

The capacitance of a capacitor is $4 \times 10^{-6} \ F$ and it is charged to a potential difference of $100 \ V$. The energy required to completely discharge it is . . . . . . $J$.

A
$0.02$
B
$0.04$
C
$0.05$
D
$0.025$

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Class 12

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Electric Potential and Capacitance

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Energy Stored in a Capacitor

The mean electrical energy density between the plates of a charged air capacitor is (where $q=$ charge on capacitor,$A=$ area of capacitor plate):

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What is the average electric energy density between the plates of a charged capacitor? (Here,$q$ = charge on the capacitor and $A$ = area of the capacitor plates.)

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$A$ capacitor is charged by a battery and the energy stored in it is $U$. The battery is now removed and the separation distance between the plates is doubled. The energy stored now is:

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If the distance between the plates of a capacitor with capacitance $C$ and charge $Q$ is doubled,the work done is:

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The amount of work done in increasing the voltage across the plates of a capacitor from $5 V$ to $10 V$ is $W$. The work done in increasing it from $10 V$ to $15 V$ will be

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The capacitance of a capacitor is $4 \times 10^{-6} \ F$ and it is charged to a potential difference of $100 \ V$. The energy required to completely discharge it is . . . . . . $J$.

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