Question

A cube of side $0.1 \, \text{m$ is placed, as shown in the figure, in a region where electric field $\vec{E} = 500x \, \hat{i}$ exists. Here $x$ is in meters and $E$ in $\text{N/C}$. Calculate:}
\textbf{(a) the flux passing through the cube, and

Hint:

For calculating flux through a cube with a non-uniform field, consider the electric field at each relevant face and use Gauss's law to find the enclosed charge.

Answer 1

\normalfont The electric flux $\Phi$ through a surface is given by: \[ \Phi = \int \vec{E} \cdot d\vec{A}, \] where: \begin{itemize} \item $\vec{E}$ is the electric field, \item $d\vec{A}$ is the area vector perpendicular to the surface. \end{itemize} % Option (a) The flux through the cube:
The electric field varies as $\vec{E} = 500x \, \hat{i}$. Only the two faces of the cube perpendicular to the $x$-axis contribute to the flux. Let the cube extend from $x = 0$ to $x = 0.1 \, \text{m}$. At $x = 0$: The electric field is: \[ \vec{E} = 500 \times 0 = 0 \, \text{N/C}. \] At $x = 0.1 \, \text{m$:} The electric field is: \[ \vec{E} = 500 \times 0.1 = 50 \, \text{N/C}. \] The area of each face of the cube is: \[ A = (0.1)^2 = 0.01 \, \text{m}^2. \] The flux through the face at $x = 0$ is: \[ \Phi_1 = \vec{E} \cdot A = 0 \cdot 0.01 = 0 \, \text{Nm}^2/\text{C}. \] The flux through the face at $x = 0.1 \, \text{m}$ is: \[ \Phi_2 = \vec{E} \cdot A = 50 \cdot 0.01 = 0.5 \, \text{Nm}^2/\text{C}. \] The net flux through the cube is: \[ \Phi = \Phi_2 - \Phi_1 = 0.5 - 0 = 0.5 \, \text{Nm}^2/\text{C}. \] Thus, the flux passing through the cube is: \[ \boxed{\Phi = 0.5 \, \text{Nm}^2/\text{C}}. \] % Option (b) The charge within the cube:
Using Gauss's law: \[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}, \] where $\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{Nm}^2$ is the permittivity of free space. Rearrange to find $q_{\text{enclosed}}$: \[ q_{\text{enclosed}} = \Phi \cdot \epsilon_0. \] Substitute the values: \[ q_{\text{enclosed}} = 0.5 \cdot 8.854 \times 10^{-12} = 4.427 \times 10^{-12} \, \text{C}. \] Thus, the charge within the cube is: \[ \boxed{q_{\text{enclosed}} = 4.43 \times 10^{-12} \, \text{C}}. \]