Question

A charge q is placed at the centre of the open end of a cylindrical vessel as shown in figure. The flux of the electric field through the surface of the vessel is :

• A. zero
• B. $\frac{q}{{\epsilon }_0}$
• C. $\frac{q}{2{\epsilon }_0}$
• D. $\frac{2q}{{\epsilon }_0}$

Answer 1

The correct option is C $\frac{q}{2{\epsilon }_0}$

As the charge $q$ is placed at the center of open end of vessel, so only charge $\frac{q}{2}$ will remain inside the vessel.

Using Gauss's law, the electric flux through vessel surface is , $\varphi =\frac{Q_{enclosed}}{{ϵ}_0}$ where $Q_{enclosed}$ is the charge inside the vessel, i.e, $Q_{enclosed}=\frac{q}{2}$

thus, $\varphi =\frac{q}{2{ϵ}_0}$