Question

The total charge contained within the cube (see figure), in which the electric field is given by \( \vec{E} = K(4x^2 \hat{i} + 3y \hat{j}) \), where \( \varepsilon_0 \) is the permittivity of free space, is: 

• A. \( 7K\varepsilon_0 \)
• B. \( 5K\varepsilon_0 \)
• C. \( 3K\varepsilon_0 \)
• D. Zero

Hint:

When using Gauss’s law in differential form, always compute \( \nabla \cdot \vec{E} \) and integrate over the entire volume to find total charge.

Answer 1

The Correct Option is A

Step 1: Apply Gauss’s law in differential form.
According to Gauss’s law, \[ \rho = \varepsilon_0 (\nabla \cdot \vec{E}). \] Step 2: Compute divergence of the given electric field.
Given \( \vec{E} = K(4x^2 \hat{i} + 3y \hat{j}) \), \[ \nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} = K(8x + 3). \] Step 3: Determine total charge in the cube.
Total charge \( Q = \int \rho\, dV = \varepsilon_0 \int (\nabla \cdot \vec{E})\, dV. \) Since the cube extends from \( x = 0 \) to \( 1 \), \( y = 0 \) to \( 1 \), and \( z = 0 \) to \( 1 \): \[ Q = \varepsilon_0 K \int_0^1 \int_0^1 \int_0^1 (8x + 3)\, dx\, dy\, dz. \] Step 4: Integrate.
\[ Q = \varepsilon_0 K \left[ 4x^2 + 3x \right]_0^1 = \varepsilon_0 K (4 + 3) = 7K\varepsilon_0. \] Step 5: Final Answer.
Total charge within the cube is \( 7K\varepsilon_0. \)