Question:
An electric field \(\overrightarrow{E} = 4x\,\hat{i} - (y^2 + 1)\hat{j}\) N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as \(\phi_I\) and \(\phi_{II}\) respectively. The difference between \((\phi_I - \phi_{II})\) is (in Nm\(^2\)/C) _______.
An electric field \(\overrightarrow{E} = 4x\,\hat{i} - (y^2 + 1)\hat{j}\) N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as \(\phi_I\) and \(\phi_{II}\) respectively. The difference between \((\phi_I - \phi_{II})\) is (in Nm\(^2\)/C) _______.
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ABCD is the top face (z = 2) and BCGF is the face x = 3; find the component of E along each outward normal and multiply by the area.
Updated On: Nov 5, 2026
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Correct Answer: -48
Solution and Explanation
Coordinates: A(0,0,2), B(3,0,2), C(3,2,2), D(0,2,2) so ABCD is the face \(z = 2\) with normal along \(\hat{k}\). \(\vec{E}\) has no z-component, so \(\phi_I = 0\).
B(3,0,2), C(3,2,2), G(3,2,0), F(3,0,0) so BCGF is the face \(x = 3\) (area \(2\times 2 = 4\) m\(^2\)) with outward normal along \(\hat{i}\). On this face \(E_x = 4(3) = 12\) N/C, so \(\phi_{II} = 12\times 4 = 48\) Nm\(^2\)/C.
\(\phi_I - \phi_{II} = 0 - 48 = -48\) Nm\(^2\)/C.
Answer: -48
B(3,0,2), C(3,2,2), G(3,2,0), F(3,0,0) so BCGF is the face \(x = 3\) (area \(2\times 2 = 4\) m\(^2\)) with outward normal along \(\hat{i}\). On this face \(E_x = 4(3) = 12\) N/C, so \(\phi_{II} = 12\times 4 = 48\) Nm\(^2\)/C.
\(\phi_I - \phi_{II} = 0 - 48 = -48\) Nm\(^2\)/C.
Answer: -48
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