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

The electric flux (Φ) through a closed surface such as a cube can be calculated using Gauss's law: \( Φ = \oint \vec{E} \cdot d\vec{A} \), where \( \vec{E} \) is the electric field, and \( d\vec{A} \) is the differential area vector. Given the electric field \( \vec{E} = (2x \hat{i}) \, \text{N C}^{-1} \), the flux through surfaces perpendicular to other axes will be zero because the field is only in the x-direction.

Consider the cube positioned from \( x = 0 \) to \( x = 2 \) with side length \( a = 2 \, \text{m} \). The cube has two faces perpendicular to the x-axis at \( x = 0 \) and \( x = 2 \). The area \( A \) of each face is \( A = 2 \times 2 = 4 \, \text{m}^2 \).

For the face at \( x = 0 \):

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

For the face at \( x = 2 \):

\( \vec{E} = 2(2) \hat{i} = 4 \, \text{N C}^{-1} \) and outward \( d\vec{A} = 4 \hat{i} \), so flux, \( Φ_2 = E \cdot A = 4 \times 4 = 16 \, \text{N m}^2/\text{C} \).

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

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

Answer 2

The electric flux is given by Gauss's Law: 
\(\Phi = \oint \vec{E} \cdot d\vec{A}.\) 
The field $\vec{E} = 2x \hat{i}$ varies with $x$. 
For the cube, only the left ($x = 0$) and right ($x = 2$) faces contribute: \(\Phi = E_\text{right} A - E_\text{left} A.\)
Substituting $A = 4 \, \mathrm{m}^2$, $E_\text{right} = 2(2) = 4$, and $E_\text{left} = 2(0) = 0$:
 \(\Phi = 4 \cdot 4 - 0 \cdot 4 = 16 \, \mathrm{Nm}^2/\mathrm{C}.\)