Question

A non-conducting cylinder is placed inside the electric field (E). The axis of the cylinder is parallel to the electric field. How much will be the total electric flux passing through the cylinder?

Hint:

No charge is enclosed, so by Gauss's law the net flux is zero; equivalently the flux entering one flat face equals the flux leaving the other.

Answer 1

Step 1: Concept.
Electric flux through a surface is \( \Phi = \vec{E}\cdot\vec{A} = EA\cos\theta \), where \( \theta \) is the angle between the field \( \vec{E} \) and the outward area vector \( \vec{A} \). The cylinder has three parts: the two flat end faces and the curved side.

Step 2: Flux through the two flat faces.
Let each flat face have area \( A \). The axis is parallel to \( \vec{E} \).
At the face where field lines enter, the outward normal points opposite to \( \vec{E} \), so \( \theta = 180^\circ \):
\[ \Phi_1 = EA\cos 180^\circ = -EA \]
At the face where field lines leave, the outward normal is along \( \vec{E} \), so \( \theta = 0^\circ \):
\[ \Phi_2 = EA\cos 0^\circ = +EA \]

Step 3: Flux through the curved surface.
On the curved side the outward normal is perpendicular to the axis, hence perpendicular to \( \vec{E} \), so \( \theta = 90^\circ \):
\[ \Phi_3 = EA'\cos 90^\circ = 0 \]

Step 4: Total flux.
\[ \Phi = \Phi_1 + \Phi_2 + \Phi_3 = -EA + EA + 0 = 0 \]
This also agrees with Gauss's law: the cylinder encloses no charge, so \( \Phi = q_{enc}/\varepsilon_0 = 0 \).

\[\boxed{\Phi_{total} = 0}\]