Question

The radius of the gaussian surface enclosing in a charge $q$ is halved, then the electric flux through the gaussian surface becomes:
A. Half
B. double
C. no change
D. infinite

Answer 1

Hint: According to Gauss law, the flux of the electric field through the closed surface is called Gaussian surface. The Gaussian surface is derived as the enclosed net charge by the permittivity of free space. It doesn’t depend on the size, shape, and position of the closed surface. Either small or big there is no change of electric flux in the Gaussian surface.

Complete step by step answer:
The electric surface through enclosed Gaussian surface remains same; there is no change in the electric surface.Therefore Gauss law,
${\varphi _{\text{closed surface}}} = \dfrac{{{q_{enc}}}}{{{e_ \circ }}}$
Closed surface of Gaussian law is equal to the enclosed surface of net charge by permittivity of the free space.As it doesn't depend upon the radius of the gaussian surface therefore the electric flux through the gaussian surface remains constant.

Hence, the correct answer is option C.

Note: Gaussian surface remains same when the charge is enclosed there is no change in the electric flux through Gaussian surface. Torque at magnet field the angle is ${90^ \circ }$. We calculated the above equation as per gauss law, torque is calculated from gauss law.