Question

An electric field \( \vec{E} = (2x \hat{i}) \, \text{N C}^{-1} \) exists in space. A cube of side \( 2 \, \text{m} \) is placed in the space as per the figure given below. The electric flux through the cube is __________ \( \text{N m}^2/\text{C} \).

Answer 1

Correct Answer: 16

Gauss's law, \( Φ = \oint \vec{E} \cdot d\vec{A} \), quantifies the electric flux \( Φ \) through a closed surface like a cube, where \( \vec{E} \) is the electric field and \( d\vec{A} \) is the differential area vector. Given \( \vec{E} = (2x \hat{i}) \, \text{N C}^{-1} \), flux through surfaces not perpendicular to the x-axis is zero due to the field's unidirectional nature.

Consider a cube spanning \( x = 0 \) to \( x = 2 \) with side length \( a = 2 \, \text{m} \). The two faces normal to the x-axis are at \( x = 0 \) and \( x = 2 \). Each face has an area \( A = 2 \times 2 = 4 \, \text{m}^2 \).

At the face \( x = 0 \):

\( \vec{E} = 2(0) \hat{i} = 0 \). Therefore, the flux \( Φ_0 = E \cdot A = 0 \times 4 = 0 \, \text{N m}^2/\text{C} \).

At the face \( x = 2 \):

\( \vec{E} = 2(2) \hat{i} = 4 \, \text{N C}^{-1} \). The outward area vector is \( d\vec{A} = 4 \hat{i} \). Thus, the flux \( Φ_2 = E \cdot A = 4 \times 4 = 16 \, \text{N m}^2/\text{C} \).

The total flux through the cube is the sum of the fluxes through these two faces: \( Φ_{\text{total}} = Φ_0 + Φ_2 = 0 + 16 = 16 \, \text{N m}^2/\text{C} \).

The calculated total flux, \( 16 \, \text{N m}^2/\text{C} \), falls within the expected range of (16,16).