Question

An electric field \( \vec{E} \) is given by:
\[ \vec{E} = \begin{cases} +100\, \hat{i} \dfrac{\text{N}}{\text{C}} & \text{for } x>0 \\ -100\, \hat{i} \dfrac{\text{N}}{\text{C}} & \text{for } x<0 \end{cases} \] A right circular cylinder of length \( 10\, \text{cm} \) and radius \( 2\, \text{cm} \), is placed such that its axis coincides with the x-axis and its two faces are at \( x = -5\, \text{cm} \) and \( x = 5\, \text{cm} \). Calculate: (a) the net outward flux through the cylinder, and (b) the net charge inside the cylinder.

Hint:

When electric field vectors on both ends of a symmetrical Gaussian surface are equal in magnitude and opposite in direction, the net flux through the surface is zero — which implies zero net enclosed charge.

Answer 1

Step 1: Area of each circular face Radius \( r = 2\, \text{cm} = 0.02\, \text{m} \) \[ A = \pi r^2 = \pi \times (0.02)^2 = 1.256 \times 10^{-3}\, \text{m}^2 \] Step 2: Electric flux through the left and right faces: Electric field at \( x = -5\, \text{cm} \): \( \vec{E}_1 = -100\, \hat{i} \) Electric field at \( x = 5\, \text{cm} \): \( \vec{E}_2 = +100\, \hat{i} \) Flux through left face (inward): \[ \phi_1 = \vec{E}_1 \cdot \vec{A} = -100 \times A = -100 \times 1.256 \times 10^{-3} = -0.1256\, \text{Nm}^2/\text{C} \] Flux through right face (outward): \[ \phi_2 = \vec{E}_2 \cdot \vec{A} = +100 \times A = +0.1256\, \text{Nm}^2/\text{C} \] Step 3: Net outward flux \[ \phi_{\text{net}} = \phi_2 + \phi_1 = 0.1256 + (-0.1256) = 0 \] Step 4: Using Gauss's Law to find net charge inside the cylinder: \[ \phi = \dfrac{q_{\text{in}}}{\varepsilon_0} \Rightarrow q_{\text{in}} = \phi_{\text{net}} \times \varepsilon_0 = 0 \times 8.85 \times 10^{-12} = 0 \] Final Answers:
(a) Net outward flux = \( 0 \, \text{Nm}^2/\text{C} \)
(b) Net charge inside the cylinder = \( 0 \, \text{C} \)