Question

A closed cylinder of radius $R$ and height $3R$ is tilted by angle $45^\circ$ with the vertical and placed in uniform horizontal electric field of magnitude $E$. (Take $\pi=\frac{22}{7}$). The net outgoing flux through the cylinder is $\frac{\alpha \sqrt{2}R^2E}{7}$ unit. Find $\alpha$.

Answer 1

0

Answer 2

According to Gauss's Law, the net electric flux through any closed surface is directly proportional to the total electric charge enclosed within that surface. In this problem, the electric field is uniform, and there is no charge mentioned to be enclosed within the closed cylinder. Therefore, the total charge enclosed ($Q_{enc}$) is zero. Consequently, the net electric flux through the closed cylinder must also be zero. Equating the given expression for the net flux, $\frac{\alpha \sqrt{2}R^2E}{7}$, to zero yields $\alpha = 0$. The dimensions and orientation of the cylinder are irrelevant for the net flux calculation in a uniform electric field.