Question

A cubical volume is bounded by the surfaces \( x = 0 \), \( x = a \), \( y = 0 \), \( y = a \), \( z = 0 \), \( z = a \). The electric field in the region is given by: \[ \vec{E} = E_0 x \hat{i} \]
Where \( E_0 = 4 \times 10^4 \, \text{NC}^{-1} \, \text{m}^{-1} \). If \( a = 2 \, \text{cm} \), the charge contained in the cubical volume is \( Q \times 10^{-14} \, \text{C} \). The value of \( Q \) is ______. (Take \( \epsilon_0 = 9 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))

Hint:

Remember Gauss’s law and the formula for electric flux. Pay close attention to the direction of the electric field and the area vector when calculating the flux.

Answer 1

Correct Answer: 288

Step 1: Visualize the Cube and Electric Field
The electric field is in the x-direction and its magnitude varies with x. The cube has side length \(a\).

Step 2: Calculate the Electric Flux
The electric flux through a surface is given by: \[ \Phi = \int \vec{E} \cdot \vec{A} \] Since the electric field is only in the x-direction, the only flux is through the face of the cube at \(x = a\) (the face ABCD in the given figure). The electric field at this face is \(\vec{E} = E_0a \hat{i}\), and the area vector is \(a^2 \hat{i}\). Therefore, the net flux through the cube is: \[ \Phi_{\text{net}} = \Phi_{\text{ABCD}} = E_0a \cdot a^2 = E_0a^3 \]

Step 3: Apply Gauss’s Law
Gauss’s law states that the net electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (\(\epsilon_0\)): \[ \Phi_{\text{net}} = \frac{q_{\text{en}}}{\epsilon_0} \] where \(q_{\text{en}}\) is the enclosed charge.

Step 4: Calculate the Enclosed Charge
Combining the flux calculation and Gauss’s law, we have: \[ \frac{q_{\text{en}}}{\epsilon_0} = E_0a^3 \] \[ q_{\text{en}} = E_0\epsilon_0a^3 \] Substitute the given values, \(E_0 = 4 \times 10^4 \, \text{N/Cm}\), \(a = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}\), and \(\epsilon_0 = 9 \times 10^{-12} \, \text{C}^2/\text{Nm}^2\): \[ q_{\text{en}} = (4 \times 10^4) \times (9 \times 10^{-12}) \times (2 \times 10^{-2})^3 \] \[ q_{\text{en}} = 36 \times 10^{-8} \times 8 \times 10^{-6} = 288 \times 10^{-14} \, \text{C} \]

Step 5: Find the Value of Q
The enclosed charge is given as \(Q \times 10^{-14} \, \text{C}\). We have found \(q_{\text{en}} = 288 \times 10^{-14} \, \text{C}\). Therefore, \(Q = 288\).

Conclusion: The value of \(Q\) is 288.

Answer 2

The correct answer is 288.

$\vec{E}=E_0\times \hat{i}$
${\varphi }_{\text{net}}={\varphi }_{ABCD}=E_{0}a\cdot a^2$
$\frac{q_{\text{en}}}{{ϵ}_0}=E_0{a}^3$
$q_{en}=E_0{\in }_0{a}^3$
$=4\times 10^4\times 9\times 10^{-12}\times 8\times 10^{-6}$
$=288\times 10^{-14}C$
$Q=288$