Question

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

• A. $\frac{2q}{{\in }_0}$
• B. $\frac{q}{2{\in }_0}$
• C. Zero
• D. $\frac{q}{{\in }_0}$

Answer 1

The correct option is B $\frac{q}{2{\in }_0}$

To close the charge $q$, we can assume an imaginary Gaussian surface of same shape as of cylindrical vessel and kept over cylindrical vessel as shown below.


Now, by Gauss's law, electric flux through closed surface is,

$\varphi =\frac{q}{{\in }_0}$

Further, the electric flux through the cylindrical vessel will be half of the total flux (by symmetry)

${\varphi }_0=\frac{\varphi }{2}=\frac{q}{2{\in }_0}$