Question

Due to a charge inside a cube the electric field is Ex = 600 x1/2, Ey = 0, Ez = 0. The charge inside the cube is (approximately)

Answer 1

7 x 10^-12 C

Answer 2

To determine the approximate charge inside the cube, we apply Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space:

$\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$

Given the electric field components $E_x = 600x^{1/2}$, $E_y = 0$, and $E_z = 0$, we consider a cube with side length $a = 0.1 \, \text{m}$ extending from $x = 0.1 \, \text{m}$ to $x = 0.2 \, \text{m}$. The area of each face is $A = a^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2$.

1. Flux through the face at $x = 0.1 \, \text{m}$ (Left face):

   The electric field at this face is $\vec{E}_{0.1} = 600(0.1)^{1/2} \hat{i}$. The area vector is $d\vec{A} = dA(-\hat{i})$.

   The flux is $\Phi_{0.1} = \int \vec{E}_{0.1} \cdot d\vec{A} = -600(0.1)^{1/2}A = -600 \times (0.1)^{1/2} \times 0.01 \approx -1.897 \, \text{Nm}^2/\text{C}$.

2. Flux through the face at $x = 0.2 \, \text{m}$ (Right face):

   The electric field at this face is $\vec{E}_{0.2} = 600(0.2)^{1/2} \hat{i}$. The area vector is $d\vec{A} = dA(+\hat{i})$.

   The flux is $\Phi_{0.2} = \int \vec{E}_{0.2} \cdot d\vec{A} = 600(0.2)^{1/2}A = 600 \times (0.2)^{1/2} \times 0.01 \approx 2.683 \, \text{Nm}^2/\text{C}$.

The total electric flux through the cube is:

$\Phi = \Phi_{0.1} + \Phi_{0.2} = -1.897 + 2.683 = 0.786 \, \text{Nm}^2/\text{C}$.

Using Gauss's Law, $q_{enclosed} = \Phi \epsilon_0$, where $\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{Nm}^2$:

$q_{enclosed} = (0.786 \, \text{Nm}^2/\text{C}) \times (8.854 \times 10^{-12} \, \text{C}^2/\text{Nm}^2) \approx 6.958 \times 10^{-12} \, \text{C}$.

Therefore, the approximate charge inside the cube is $7 \times 10^{-12} \, \text{C}$.