(A) Given: Capacitance $C = 1 \ \mu F = 10^{-6} \ F$,Potential difference $V = 20 \ V$,Distance $d = 1 \ \mu m = 10^{-6} \ m$.The electric field $E$ b

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(A) Given: Capacitance $C = 1 \ \mu F = 10^{-6} \ F$,Potential difference $V = 20 \ V$,Distance $d = 1 \ \mu m = 10^{-6} \ m$.The electric field $E$ between the plates is given by $E = \frac{V}{d} = \frac{20}{10^{-6}} = 20 \times 10^6 \ V/m$.The energy density $u$ is given by the formula $u = \frac{1}{2} \epsilon_0 E^2$.Substituting the values: $u = \frac{1}{2} \times (8.854 \times 10^{-12}) \times (20 \times 10^6)^2$.$u = 0.5 \times 8.854 \times 10^{-12} \times 400 \times 10^{12}$.$u = 0.5 \times 8.854 \times 400 = 1770.8 \ J/m^3 \approx 1.8 \times 10^3 \ J/m^3$.

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

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$A$ parallel plate capacitor of capacitance $1 \ \mu F$ is charged to a potential difference of $20 \ V$. The distance between the plates is $1 \ \mu m$. The energy density between the plates of the capacitor is:

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A
$1.8 \times 10^3 \ J/m^3$
B
$2 \times 10^{-4} \ J/m^3$
C
$2 \times 10^2 \ J/m^3$
D
$1.8 \times 10^5 \ J/m^3$

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Electric Potential and Capacitance

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Energy Stored in a Capacitor

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$A$ parallel plate capacitor of capacitance $1 \ \mu F$ is charged to a potential difference of $20 \ V$. The distance between the plates is $1 \ \mu m$. The energy density between the plates of the capacitor is:

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$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the energy stored in the capacitor.

In the figure,the charge on the capacitor is plotted against the potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively:

$A$ parallel plate capacitor has a uniform electric field $E$ 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:

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