An electric field overrightarrow{E} = 4x hat{ | vedclass

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(D) The electric field is given by $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j}$.Surface $ABCD$ lies in the $xy$-plane at $z = 2$. The area ve

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$-48$

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(D) The electric field is given by $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j}$.Surface $ABCD$ lies in the $xy$-plane at $z = 2$. The area vector for this surface is $\overrightarrow{S}_I = S \hat{k}$. Since $\overrightarrow{E} \cdot \hat{k} = 0$,the flux $\phi_I = 0$.Surface $BCGF$ lies in the $yz$-plane at $x = 3$. The area vector for this surface is $\overrightarrow{S}_{II} = 4 \hat{i}$ (area = $2 	imes 2 = 4$).At $x = 3$,the electric field is $\overrightarrow{E} = 4(3) \hat{i} - (y^2 + 1) \hat{j} = 12 \hat{i} - (y^2 + 1) \hat{j}$.The flux $\phi_{II} = \int \overrightarrow{E} \cdot d\overrightarrow{S} = \int_{0}^{2} \int_{0}^{2} (12 \hat{i} - (y^2 + 1) \hat{j}) \cdot (dy dz \hat{i}) = \int_{0}^{2} \int_{0}^{2} 12 \, dy dz = 12 	imes 4 = 48 	ext{ Nm}^2/C$.Thus,$\phi_I - \phi_{II} = 0 - 48 = -48 	ext{ Nm}^2/C$.

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