Example

Question:

The electric field components in Fig. 1.24 are \( E_x = \alpha x^{1/2} \), \( E_y = E_z = 0 \), in which \( \alpha = 800\,\text{N}\,\text{C}^{-1}\,\text{m}^{1/2} \). Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that \( a = 0.1\,\text{m} \).

Solution:

(a) Since the electric field has only an \( x \) component, for faces perpendicular to the \( x \) direction, the angle between \( \mathbf{E} \) and \( \Delta \mathbf{S} \) is \( \pm \pi/2 \). Therefore, the flux \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} \) is zero for each face except the two shaded ones.
Magnitude of the electric field at the left face: \[ E_L = \alpha x^{1/2} = \alpha a^{1/2} \] Magnitude at the right face: \[ E_R = \alpha x^{1/2} = \alpha (2a)^{1/2} \] The corresponding fluxes are: \[ \phi_L = E_L (-a^2),\quad \phi_R = E_R a^2 \] Net flux through the cube: \[ \phi_R + \phi_L = E_R a^2 - E_L a^2 = a^2 (E_R - E_L) \] \[ = \alpha a^{5/2} (\sqrt{2} - 1) \] \[ = 800\, (0.1)^{5/2} (\sqrt{2} - 1) = 1.05\,\text{N}\,\text{m}^2\,\text{C}^{-1} \] (b) Using Gauss's law, \( \phi = q / \varepsilon_0 \) or \( q = \phi \varepsilon_0 \): \[ q = 1.05 \times 8.854 \times 10^{-12}\,\text{C} = 9.27 \times 10^{-12}\,\text{C} \]