Question

Choose the correct option:
The value of distance $ {r_m} $ at which electric field intensity is maximum, is given by:
(A) $ {r_m} = \dfrac{r}{3} $
(B) $ {r_m} = \dfrac{{3R}}{2} $
(C) $ {r_m} = \dfrac{{2R}}{3} $ $ $
(D) $ {r_m} = \dfrac{{4R}}{3} $

Answer 1

Hint
Using Gauss’ law, the field at a distance $ r $ from the center of the sphere is $ E.4\pi {r^2} = \dfrac{{{Q_{en}}}}{\varepsilon } $ . Here, $ {Q_{en}} $ is the charge enclosed by the gaussian surface, and the $ \varepsilon $ is the permittivity of the medium. Also, for calculating maximum field intensity, $ \dfrac{{dE}}{{dr}} = 0 $ .

Complete step by step answer
The intensity or strength of an electric field at any point in the field is the force experienced by a unit positive charge placed at that point.
Now, using Gauss’ law, the field at a distance $ r $ from the center of the sphere is $ E.4\pi {r^2} = \dfrac{{{Q_{en}}}}{\varepsilon } $ (where, $ {Q_{en}} $ is the charge enclosed by the gaussian surface, and the $ \varepsilon $ is the permittivity of the medium).
Now, $ {Q_{en}} = {\smallint _0}^r\rho (4\pi {r^2})dr = 4\pi {\rho _0}{\smallint _0}^r[1 - r/R]{r^2}dr = 4\pi {\rho _0}{[{r^3}/3 - {r^4}/4R]_0}^r = 4\pi {\rho _0}[{r^3}/3 - {r^4}/4R] $
So, $ E.4\pi {r^2} = \dfrac{{4\pi {\rho _0}[{r^3}/3 - {r^4}/4R]}}{\varepsilon } $
 $ or,E = \dfrac{{{\rho _0}}}{\varepsilon }[r/3 - {r^2}/4R] $
For calculating maximum field intensity, $ \dfrac{{dE}}{{dr}} = 0 = \dfrac{{{\rho _0}}}{\varepsilon }[\dfrac{1}{3} - \dfrac{{2r}}{{4R}}] $
or, $ \dfrac{1}{3} = \dfrac{{2r}}{{4R}} = \dfrac{r}{{2R}} $
or, $ r = {r_m} = \dfrac{{2R}}{3} $.
Option (C) is the correct answer.

Additional Information
The Gauss’ law states that the net electric flux linked with a closed surface is $ \dfrac{1}{\varepsilon } $ times the net charge within the surface.
Mathematically, $ \phi = {\smallint _{closed}}E.ds = \dfrac{{{Q_{en}}}}{\varepsilon } $
Only the charges inside a volume contribute to the electric flux linked with the surface enclosing that volume. If the closed surface encloses equal amounts of positive and negative charges, then the net charge is zero; consequently, the flux linked with the surface is also zero. And, also, charges outside the closed surface have no net contribution towards the electric flux linked with it.

Note
Obviously the intensity of an electric field is a vector quantity. The direction of intensity is given by the direction of force acting on the positive charge. An electric field may be generated due to more than one charge. Intensity at a point in such an electric field is determined by calculating the field at that point due to all the charges individually and then the resultant of all the field strengths are found out by vector addition.