A parallel plate capacitor of capacitance 1 | vedclass

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Question

(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

Vedclass · Accepted Answer

$1.8 	imes 10^3 \ J/m^3$

Vedclass · Answer

(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 $E = \frac{V}{d} = \frac{20}{10^{-6}} = 20 	imes 10^6 \ V/m$.The energy density $u$ is given by the formula $u = \frac{1}{2} \epsilon_0 E^2$.Substituting the values: $u = \frac{1}{2} 	imes (8.854 	imes 10^{-12}) 	imes (20 	imes 10^6)^2$.$u = 0.5 	imes 8.854 	imes 10^{-12} 	imes 400 	imes 10^{12}$.$u = 0.5 	imes 8.854 	imes 400 = 1770.8 \ J/m^3 \approx 1.8 	imes 10^3 \ J/m^3$.

$A$ 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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