Question

A charge \[Q\] is enclosed by a Gaussian spherical surface of radius \[R\]. If the radius is doubled, then the outward electric flux will:
A. be doubled
B. increase four times
C. be reduced to half
D. remains the same

Answer 1

Hint: The net electric flux coming out of the Gaussian spherical surface depends on the nature as well as the quantity of the charges enclosed by that surface. It also depends on the medium present. But, it is independent of the distribution of the charges and the separation between then inside the closed surface. Thus, it is independent of the radius of the Gaussian sphere.

Formula Used:
The Gauss Law is given as:
\[\phi = \oint\limits_s {\overrightarrow E .\overrightarrow {dS} } = \dfrac{q}{{{\varepsilon _ \circ }}}\]
where, \[\overrightarrow E \] is the electric field intensity, \[\overrightarrow {dS} \] is a small surface element, \[q\] is the charge and \[{\varepsilon _0}\] is the permittivity in free space.

Complete step by step answer:
The imaginary sphere enclosing the position or distribution of charges is defined as Gaussian surface. The total number of lines of force that pass through a closed surface in an electric field is called electric flux. It is denoted as \[\phi \]. According, to Gauss Law, the net electric flux through any closed surface is equal to \[\dfrac{1}{{{\varepsilon _0}}}\]times the total electric charge enclosed by the surface. That is,
\[\phi = \oint\limits_s {\overrightarrow E .\overrightarrow {dS} } = \dfrac{q}{{{\varepsilon _ \circ }}}\]
Thus, the net electric flux through any closed spherical Gaussian surface depends on the factors \[q\] and \[{\varepsilon _0}\] and does not depend on the radius of the Gaussian sphere. Even if the radius is doubled, then the outward electric flux will remain the same as \[\dfrac{q}{{{\varepsilon _0}}}\]. This is because the outward electric flux is independent of the distribution of the charges and the separation between them inside the closed surface.

Hence, option D is the correct answer.

Note: The electric flux entering the closed surface is taken as negative and that emanating out of the closed surface is taken as positive.Gaussian surface is spherical for a point charge, conducting and non-conducting spheres and it is cylindrical for infinite sheet of charge and infinite line of charge.