In a uniform electric field,a cube of side 1 | vedclass

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(B) The energy density $u$ in an electric field is given by $u = \frac{1}{2} \epsilon_0 E^2$. Total energy $U = u 	imes V$,where $V$ is the volume of

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$100\sqrt{2} \ V \cdot m$

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(B) The energy density $u$ in an electric field is given by $u = \frac{1}{2} \epsilon_0 E^2$. Total energy $U = u 	imes V$,where $V$ is the volume of the cube. Given $U = 8.85 \ \mu J = 8.85 	imes 10^{-6} \ J$,side $a = 1 \ cm = 10^{-2} \ m$,so $V = a^3 = 10^{-6} \ m^3$. Using $\epsilon_0 = 8.85 	imes 10^{-12} \ C^2/(N \cdot m^2)$: $8.85 	imes 10^{-6} = \frac{1}{2} 	imes (8.85 	imes 10^{-12}) 	imes E^2 	imes 10^{-6}$. $1 = \frac{1}{2} 	imes 10^{-12} 	imes E^2 \implies E^2 = 2 	imes 10^{12} \implies E = \sqrt{2} 	imes 10^6 \ V/m$. The electric field is parallel to four faces,meaning it is perpendicular to the remaining two faces. The flux $\phi$ through one of these faces is $\phi = E 	imes A = E 	imes a^2$. $\phi = (\sqrt{2} 	imes 10^6) 	imes (10^{-2})^2 = \sqrt{2} 	imes 10^6 	imes 10^{-4} = 100\sqrt{2} \ V \cdot m$.

In a uniform electric field,a cube of side $1 \ cm$ is placed. The total energy stored in the cube is $8.85 \ \mu J$. The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is.

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