Question

A sphere of radius R and charge Q is placed inside an imaginary sphere of radius 2R whose center coincides with the given sphere. The flux related to the imaginary sphere is
A) $\dfrac{{{Q}}}{{{ \in _{{0}}}}} \\$
B) $\dfrac{{{Q}}}{{2{ \in _{{0}}}}} \\$
C) $\dfrac{{4Q}}{{{ \in _{{0}}}}} \\$
D) $\dfrac{{2Q}}{{{ \in _{{0}}}}} \\ $

Answer 1

Hint: Sphere of charge Q and having radius R is placed inside an imaginary sphere having radius double to that of the actual sphere. We can use Gauss's law to find the flux related to the imaginary sphere.

Complete step by step solution:
Electric flux: It is the measure of the number of electric field lines passing through a given surface normally. It is denoted by ${\phi _E}$. It is a scalar quantity.
Gauss theorem states that the total electric flux through a closed surface in vacuum is equal to the $\dfrac{1}{{{ \in _0}}}$times the total charge enclosed by the closed surface i.e.
${\phi _{{E}}} = \oint {\vec E.d\vec A} = \dfrac{Q}{{{ \in _0}}}$
Where Q = total charge enclosed
${ \in _0}$ (Epsilon not ) = permittivity of free space
Electrical permittivity is the measure of the resistance offered against the formation of an electric field. The value of ${ \in _0}$ = $8.85 \times {10^{-12}} {C^2}/N-m$.

The most important point regarding Gauss theorem is that the value of the electric flux i.e. ${\phi _E}$ does not get affected by the shape or size of closed surface that means from charge Q that is placed inside an imaginary cylinder of radius 2R will be $\dfrac{Q}{{{\varepsilon _0}}}$.
So, flux related to the imaginary sphere is $\dfrac{Q}{{{\varepsilon _0}}}$.

Therefore, option (A) is the correct choice.

Note: There are many drawbacks of Gauss theorem. Some of them are as follow:
1) The Gaussian surface should be symmetric about charge.
2) The electric field must be symmetric at all points of Gaussian surface.
3) Angle ${{\theta }}$ between electric field vector and area vector must be the same at all the points of the surface.