The electric field in a region of space varies as `E=(xhati+2yhatj)+3zhatk)` V/m
(a). Consider an elemental cuboid whose one vertex is at (x,y,z) and the three sides are dx, dy and dz, sides being parallel to the three co-ordinate axes. Calculate the flux of electric field through the cube.
(b). Usig the expression obtained in (a) find the charge enclosed by a spherical surface of radius r, centred at the origin.

Answer

To solve the problem step by step, we will address both parts (a) and (b) of the question.

### Part (a): Calculate the flux of the electric field through the cube.

1. **Identify the Electric Field**:
The electric field is given by:
\[ \mathbf{E} = (x \hat{i} + 2y \hat{j} + 3z \hat{k}) \, \text{V/m} \]

2. **Define the Elemental Cuboid**:
Consider an elemental cuboid with one vertex at \((x, y, z)\) and dimensions \(dx\), \(dy\), and \(dz\).

3. **Calculate the Flux through Each Face of the Cuboid**:
The electric flux \(\Phi\) through a surface is given by:
\[ \Phi = \int \mathbf{E} \cdot d\mathbf{A} \]
where \(d\mathbf{A}\) is the area vector of the surface.

- **Face 1 (x = constant)**: Area vector \(d\mathbf{A} = dy \, dz \hat{i}\)
\[ \Phi_1 = \mathbf{E} \cdot d\mathbf{A} = (x \hat{i} + 2y \hat{j} + 3z \hat{k}) \cdot (dy \, dz \hat{i}) = x \, dy \, dz \]

- **Face 2 (x + dx = constant)**: Area vector \(d\mathbf{A} = -dy \, dz \hat{i}\)
\[ \Phi_2 = -\mathbf{E} \cdot d\mathbf{A} = -(x + dx)(dy \, dz) = -(x \, dy \, dz + dx \, dy \, dz) \]

- **Face 3 (y = constant)**: Area vector \(d\mathbf{A} = dx \, dz \hat{j}\)
\[ \Phi_3 = (x \hat{i} + 2y \hat{j} + 3z \hat{k}) \cdot (dx \, dz \hat{j}) = 2y \, dx \, dz \]

- **Face 4 (y + dy = constant)**: Area vector \(d\mathbf{A} = -dx \, dz \hat{j}\)
\[ \Phi_4 = -\mathbf{E} \cdot d\mathbf{A} = -(x \hat{i} + 2(y + dy) \hat{j} + 3z \hat{k}) \cdot (-dx \, dz \hat{j}) = -2(y + dy) \, dx \, dz \]

- **Face 5 (z = constant)**: Area vector \(d\mathbf{A} = dx \, dy \hat{k}\)
\[ \Phi_5 = (x \hat{i} + 2y \hat{j} + 3z \hat{k}) \cdot (dx \, dy \hat{k}) = 3z \, dx \, dy \]

- **Face 6 (z + dz = constant)**: Area vector \(d\mathbf{A} = -dx \, dy \hat{k}\)
\[ \Phi_6 = -\mathbf{E} \cdot d\mathbf{A} = -(x \hat{i} + 2y \hat{j} + 3(z + dz) \hat{k}) \cdot (-dx \, dy \hat{k}) = -3(z + dz) \, dx \, dy \]

4. **Total Flux Calculation**:
Now, summing all the flux contributions:
\[ \Phi = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 \]
\[ \Phi = (x \, dy \, dz - (x \, dy \, dz + dx \, dy \, dz) + 2y \, dx \, dz - 2(y + dy) \, dx \, dz + 3z \, dx \, dy - 3(z + dz) \, dx \, dy) \]
Simplifying this expression leads to:
\[ \Phi = 6 \, dx \, dy \, dz \]

### Part (b): Find the charge enclosed by a spherical surface of radius \(r\) centered at the origin.

1. **Use Gauss's Law**:
Gauss's law states:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \(Q_{\text{enc}}\) is the charge enclosed and \(\epsilon_0\) is the permittivity of free space.

2. **Substituting the Flux**:
From part (a), we found that the flux through the cuboid is:
\[ \Phi = 6 \, dx \, dy \, dz \]

3. **Volume Element**:
The volume element \(dV\) of the cuboid is:
\[ dV = dx \, dy \, dz \]

4. **Charge Density**:
The charge density \(\rho\) can be expressed as:
\[ \rho = \frac{Q_{\text{enc}}}{dV} = 6 \epsilon_0 \]

5. **Total Charge Enclosed in a Sphere**:
The total charge enclosed by a sphere of radius \(r\) is:
\[ Q_{\text{enc}} = \rho \cdot V = 6 \epsilon_0 \cdot \frac{4}{3} \pi r^3 \]
Therefore,
\[ Q_{\text{enc}} = 8 \epsilon_0 \pi r^3 \]

### Final Answers:
- **(a)** The flux of the electric field through the cuboid is \(6 \, dx \, dy \, dz\).
- **(b)** The charge enclosed by a spherical surface of radius \(r\) is \(8 \epsilon_0 \pi r^3\).