Question

Consider a hypothetical situation in which charge is uniformly distributed in the whole space. If net flux passing through the surface of imaginary cube of edge $a$ is $\phi$ then flux passing through the imaginary sphere of radius $a$ will be
A. $\phi$
B. $\dfrac{{4\pi \phi }}{3}$
C. $\dfrac{{4\phi }}{3}$
D. $\dfrac{{3\pi \phi }}{4}$

Answer 1

In order to solve this question, we will use the gauss’s law of electrostatics which relates the charge and flux of electric field in space and using this, we will first formula a relation between flux and charge volume density for cube and then for sphere and later by eliminating known parameters value we will solve for flux for sphere.

Formula used:
According to gauss’s law of electrostatics,
$\phi = \dfrac{q}{{{ \in _o}}}$
where, $\phi ,q,{ \in _o}$ represents flux, charge, permittivity of free space.

Complete step by step answer:
According to the question, charge in uniformly distributed in space so let us assume $\rho$ be the charge density per unit volume in space so, for cube the net charge inside will be simply the volume of cube multiplied by charge density so, volume of cube of side a will be ${a^3}$ so, charge inside the cube will be $q = {a^3}\rho$. On putting the value in equation, $\phi = \dfrac{q}{{{ \in _o}}}$
We get,
$\phi = \dfrac{{{a^3}\rho }}{{{ \in _o}}} \to (i)$
Similarly, for sphere of radius a volume of sphere of radius a will be $\dfrac{{4\pi {a^3}}}{3}$ and charge inside the sphere will be,
$q = \dfrac{{4\pi {a^3}}}{3}\rho$

Let $\phi '$ be the flux through sphere, then using the equation $\phi = \dfrac{q}{{{ \in _o}}}$
We get,
$\phi ' = \dfrac{{\dfrac{{4\pi {a^3}}}{3}\rho }}{{{ \in _o}}} \to (ii)$
Divide the equation (i) by (ii) we get,
$\dfrac{\phi }{{\phi '}} = \dfrac{{\dfrac{{{a^3}\rho }}{{{ \in _o}}}}}{{\dfrac{{\dfrac{{4\pi {a^3}}}{3}\rho }}{{{ \in _o}}}}}$
On simplifying we get,
$\Rightarrow \dfrac{\phi }{{\phi '}} = \dfrac{3}{{4\pi }}$
$\therefore \phi ' = \dfrac{{4\pi \phi }}{3}$

Hence, the correct option is B.

Note: It should be remembered that, the charge is taken the net charge inside the closed volume through which electric flux is calculated and here charge density remains same for both cases because it was said in question that charge is uniformly distributed everywhere, if it were not we can’t use same charge density in both cases.