If the energy density in a parallel plate cap | vedclass

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(B) The energy density $u$ of an electric field is given by the formula: $u = \frac{1}{2} \epsilon_0 E^2$.Given: $u = 1.8 	imes 10^{-9} \ J/m^3$ and

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$20$

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(B) The energy density $u$ of an electric field is given by the formula: $u = \frac{1}{2} \epsilon_0 E^2$.Given: $u = 1.8 	imes 10^{-9} \ J/m^3$ and $\epsilon_0 = 9 	imes 10^{-12} \ C^2/N \cdot m^2$.Substituting the values into the formula:$1.8 	imes 10^{-9} = \frac{1}{2} 	imes (9 	imes 10^{-12}) 	imes E^2$.$1.8 	imes 10^{-9} = 4.5 	imes 10^{-12} 	imes E^2$.$E^2 = \frac{1.8 	imes 10^{-9}}{4.5 	imes 10^{-12}}$.$E^2 = \frac{1.8}{4.5} 	imes 10^3 = 0.4 	imes 1000 = 400$.$E = \sqrt{400} = 20 \ NC^{-1}$.Therefore,the correct option is $B$.

If the energy density in a parallel plate capacitor is given as $1.8 \times 10^{-9} \ J/m^3$,then the value of the electric field in the region between the plates is ....... $NC^{-1}$. (Given: $\epsilon_0 = 9 \times 10^{-12} \ C^2/N \cdot m^2$)

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