
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
163.8k+ views
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.
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
$q=\dfrac{4\pi~kr^{5}}{5}$
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
