Question

A sphere of radius R carries a charge such that its volume charge density is proportional to the square of the distance from the center. What is the ratio of the magnitude of the electric field at a distance of 2R from the center to the magnitude of the electric field at a distance of R/2 from the center (i.e. $E_{R=2R}$ / $E_{R=R/2}$)?
A. 1
B. 2
C. 4
D. 8

Answer 1

Hint: When a body of a particular volume is filled with charge evenly, then the amount of charge that a unit of a volume holds is called volume charge density. So this question can be solved by the concept of continuous charge distributions.


Formula used:
The formulas used in the given question are-
The formula used for the determination of electric field,
$\vec{E}\times\vec{A}=\ \dfrac{\int{\rho\times V}}{\varepsilon_0}$
Where ρ = volume charge density
$\vec{E}$ = Electric Field
$\vec{A}$ = Area of the sphere (4πR2)
V= Volume of the sphere (4πR2)
ε0 = permittivity constant




Complete answer:

According to the question; the volume charge density (VCD) is proportional to the square of the distance from the center of a sphere of radius R. Then ,
$\rho~\ltimes~r^{2}$
$\rho=kr^{2}$
Where, k is constant
For r > R (i.e, outside the sphere of radius R)
By Gauss’s law,
$\oint\overrightarrow{E}.d\overrightarrow{A}=\dfrac{q_{enc}}{\varepsilon_{0}}$ ……(i)
$q=k\int_{0}^{R}r^{2}4\pi~r^{2}dr$
$q=\dfrac{4\pi~R^{5}}{5}$
Substituting the value of q in equation (i), we get
$E\times(4\pi r^2)=\dfrac{k\times4\pi R^5}{5\varepsilon_0}$
$E=\dfrac{k R^5}{5\varepsilon_0~r^2}$
At r=2R, we can write
$E=\dfrac{k R^5}{5\varepsilon_0~\lgroup~2R\rgroup^2}$
This implies,
$E=\dfrac{k R^3}{20\varepsilon_0}$ …….(ii)
Now, for r$q=k\int_{0}^{r}r^{2}4\pi~r^{2}dr$
$q=\dfrac{4\pi~kr^{5}}{5}$
Substituting the value of q in equation (i), we get
$E\times(4\pi r^2)=\dfrac{k\times4\pi r^5}{5\varepsilon_0}$
Further simplifying,
$E=\dfrac{k r^3}{5\varepsilon_0}$
At r=R/2
$E=\dfrac{k\lgroup\dfrac{R}{2}\rgroup^3}{5\varepsilon_0} $
Simplifying,
$E=\dfrac{k R^3}{40\varepsilon_0}$ ……(ii)
Now, the ratio of electric field at r=2R to the r=R/2,
$\dfrac{E_{r=2R}}{E_{r=\dfrac{R}{2}}}=\dfrac{\dfrac{kR^3}{20\varepsilon_0}}{\dfrac{kR^3}{40\varepsilon_0}}$
$\dfrac{E_{r=2R}}{E_{r=\dfrac{R}{2}}}=\dfrac{40}{20}$
$ \dfrac{E_{r=2R}}{E_{r=\dfrac{R}{2}}}=2$
Thus the ratio between $\dfrac{E_{r=2R}}{E_{r=\dfrac{R}{2}}}$ is equal to 2.
Thus, Option (B) is correct


Note: An area in which the charge is uniformly distributed is called the electric field. The electric field is dependent on the distance; the farther the object the lesser it will experience the electric field. Electric field is a vector quantity, it has magnitude and direction. One also needs to understand the concept of continuous charge densities.