Example

Question:

An electric field is uniform, and in the positive \( x \) direction for positive \( x \), and uniform with the same magnitude but in the negative \( x \) direction for negative \( x \). It is given that \( E = 200\,\text{N/C} \) for \( x > 0 \) and \( E = -200\,\text{N/C} \) for \( x < 0 \). A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the \( x \)-axis, so that one face is at \( x = +10\,\text{cm} \) and the other is at \( x = -10\,\text{cm} \). What is the net outward flux through each flat face? What is the flux through the side of the cylinder? What is the net outward flux through the cylinder? What is the net charge inside the cylinder?

Solution:

(a) On the left face, \( \mathbf{E} \) and \( \Delta \mathbf{S} \) are parallel: \[ \phi_L = \mathbf{E} \cdot \Delta \mathbf{S} = -200\,\hat{i} \cdot \Delta \mathbf{S} \] For outward flux: \( +200\,\Delta S \), with \( \Delta S = \pi (0.05)^2 \) \[ \phi_L = +200 \times \pi \times (0.05)^2 = 1.57\,\text{N\,m}^2\,\text{C}^{-1} \] On the right face, \[ \phi_R = \mathbf{E} \cdot \Delta \mathbf{S} = +200\,\Delta S = 1.57\,\text{N\,m}^2\,\text{C}^{-1} \] (b) For any point on the side, \( \mathbf{E} \) is perpendicular to \( \Delta \mathbf{S} \), so \( \mathbf{E} \cdot \Delta \mathbf{S} = 0 \). The flux out of the side of the cylinder is zero.
(c) Net outward flux through the cylinder: \[ \phi = 1.57 + 1.57 = 3.14\,\text{N\,m}^2\,\text{C}^{-1} \] (d) Net charge within the cylinder, by Gauss's law: \[ q = \varepsilon_0 \phi = 3.14 \times 8.854 \times 10^{-12}\,\text{C} = 2.78 \times 10^{-11}\,\text{C} \]