Question

A sphere of radius R has a volume density of charge $\rho = kr$, where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the surface of the sphere is given by-
(${\varepsilon _0}$ = permittivity of free space)

A.)$\dfrac{{4\pi k{R^2}}}{{3{\varepsilon _0}}}$
B.)$\dfrac{{kR}}{{3{\varepsilon _0}}}$
C.)$\dfrac{{4\pi kR}}{{{\varepsilon _0}}}$
D.)$\dfrac{{k{R^2}}}{{4{\varepsilon _0}}}$

Answer 1

Hint: Gauss’ theorem typically states that- Only sources (positive charges) and sinks (negative charges) of the electrical fields enclosed by the surface may allow electrical flow to come from any closed region. Electric flows are not supported by charges outside the surface. Therefore, electric charges can serve only as electric field sources or sinks. For example , changing magnetic fields cannot act as electric field sources or sinks.

Complete answer:
Formula used: $\Rightarrow \oint {\vec E.\vec ds} = \dfrac{1}{{{\varepsilon _0}}}q$

“The total flux linked with a closed surface is $\dfrac{1}{{{\varepsilon _0}}}$ times the charge enclosed by the closed surface”.

The mathematical representation being-

$\Rightarrow \oint {\vec E.\vec ds} = \dfrac{1}{{{\varepsilon _0}}}q$

Now, we already know that $\rho = kr$ as per given in the question.

Applying the gauss’ theorem as per discussed above-
Surface area of the sphere is always calculated as $4\pi {r^2}$.

$\Rightarrow E\oint {dA = } \dfrac{Q}{{{\varepsilon _0}}} \\\ \\\ \Rightarrow E\left( {4\pi {r^2}} \right) = \dfrac{{\int \rho \times 4\pi {r^2}dr}}{{{\varepsilon _0}}} \\\ \\\ \Rightarrow E\left( {4\pi {r^2}} \right) = \dfrac{{\int {Kr} \times 4\pi {r^2}dr}}{{{\varepsilon _0}}} \\\ \\\ \Rightarrow E = \dfrac{{K{r^2}}}{{4{\varepsilon _0}}} \\\$

Here, in the question, it was given that the radius of the given sphere is R. thus, substituting this value in the above calculated equation. We get-

$\Rightarrow E = \dfrac{{K{r^2}}}{{4{\varepsilon _0}}} \\\ \\\ \Rightarrow E = \dfrac{{K{R^2}}}{{4{\varepsilon _0}}} \\\$

Hence, it is clear that option D is the correct option.

Note: The rule of Gauss is only a reaffirmation of Coulomb's Law. You can easily get back to the Coulomb law if you apply the Gauss theorem to a point charge contained by one sphere. In short, the Gauss theorem refers to electric field lines (flux) 'flow' to charges in the surface enclosed. If a surface does not contain any charges, then the net electric flux remains zero.