In a region of space the electric field in the `x`-direction and proportional to `x`i.e., `vec(E )=E_(0)xhat(i)`. Consider an imaginary cubical volume of edge a with its parallel to the axes of coordinates. The charge inside this volume will be

Options

A. Zero
B. `epsilon_(0)E_(0)a^(3)`
C. `1/epsilon_(0)E_(0)a^(3)`
D. `1/6epsilon_(0)E_(0)a^(3)`

Answer

To find the charge enclosed in a cubical volume in a region where the electric field is given by \(\vec{E} = E_0 x \hat{i}\), we can use Gauss's law, which states:

\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]

where \(\Phi_E\) is the electric flux through a closed surface, \(Q_{\text{enc}}\) is the charge enclosed by that surface, and \(\epsilon_0\) is the permittivity of free space.

### Step-by-Step Solution

1. **Identify the Geometry**:
We have a cube of edge length \(a\) aligned with the coordinate axes. The cube's vertices can be labeled based on their coordinates.

2. **Calculate the Electric Flux through Each Face of the Cube**:
The electric field varies with \(x\), so we need to calculate the flux through each face of the cube.

- **Face A (x = x₀)**: The outward normal vector is \(-\hat{i}\). The electric field at this face is \(\vec{E} = E_0 x_0 \hat{i}\). The area vector \(dA\) is \(A^2 (-\hat{i})\), where \(A = a\). Thus, the flux through this face is:
\[ \Phi_A = \vec{E} \cdot dA = E_0 x_0 A^2 (-1) = -E_0 x_0 a^2 \]

- **Face B (x = x₀ + a)**: The outward normal vector is \(\hat{i}\). The electric field at this face is \(\vec{E} = E_0 (x_0 + a) \hat{i}\). The area vector \(dA\) is \(A^2 \hat{i}\). Thus, the flux through this face is:
\[ \Phi_B = \vec{E} \cdot dA = E_0 (x_0 + a) A^2 = E_0 (x_0 + a) a^2 \]

- **Faces C, D, E, F**: The electric field has no component in the \(\hat{j}\) and \(\hat{k}\) directions, so the flux through these faces is zero:
\[ \Phi_C = \Phi_D = \Phi_E = \Phi_F = 0 \]

3. **Total Electric Flux**:
The total electric flux through the cube is the sum of the fluxes through all faces:
\[ \Phi_{\text{total}} = \Phi_A + \Phi_B + \Phi_C + \Phi_D + \Phi_E + \Phi_F \]
Substituting the values:
\[ \Phi_{\text{total}} = -E_0 x_0 a^2 + E_0 (x_0 + a) a^2 = E_0 a^2 (x_0 + a - x_0) = E_0 a^3 \]

4. **Apply Gauss's Law**:
Using Gauss's law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
We can equate the total flux to the charge enclosed:
\[ E_0 a^3 = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Rearranging gives:
\[ Q_{\text{enc}} = E_0 a^3 \epsilon_0 \]

### Final Answer:
The charge enclosed in the cubical volume is:
\[ Q_{\text{enc}} = E_0 a^3 \epsilon_0 \]