
Two concentric spheres of radii $\mathrm{R}$ and $\mathrm{r}$ have positive charges $\mathrm{q}_{1}$ and $\mathrm{q}_{2}$ with equal surface charge densities. What is the electric potential at their common centre?
A) $\dfrac{\sigma}{\varepsilon_{0}}(\mathrm{R}+\mathrm{r})$
B) $\dfrac{\sigma}{\varepsilon_{0}}(\mathrm{R}-\mathrm{r})$
C) $\dfrac{\sigma}{\varepsilon_{0}}\left(\dfrac{1}{\mathrm{R}}+\dfrac{1}{\mathrm{r}}\right)$
D) $\dfrac{\sigma}{\varepsilon_{0}}\left(\dfrac{\mathrm{R}}{\mathrm{r}}\right)$
Answer
233.1k+ views
Hint: We know that by surface charge density we mean the amount of electric charge that is present on the surface. The unit to measure this is charge per unit area. The volume and the surface area of the concerned surface is taken into consideration. For this question, the surface area of the sphere is required.
Complete step by step answer:
Let us draw the diagram, for a better understanding.

Let us consider that the radius of the outer sphere is R and the radius of the inner sphere is r. Thus, we can write that:
Charge on the outer sphere, $\mathrm{q}_{1}=4 \pi \mathrm{R}^{2} \sigma$ and,
Charge on the inner sphere, $\mathrm{q}_{2}=4 \pi \mathrm{r}^{2} \sigma$
Now let us consider that the electric potential is defined as v. So, the electric potential of the common centre will be given as the summation of the charges on both the outer and the inner spheres.
Hence, we can write that:
$\Rightarrow \mathrm{v}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\mathrm{q}_{1}}{\mathrm{R}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\mathrm{q}_{2}}{\mathrm{r}}$
After the summation we get the value of v as:
$\Rightarrow \dfrac{1}{4 \pi \varepsilon_{0}}\left[\dfrac{4 \pi \mathrm{R}^{2} \sigma}{\mathrm{R}}+\dfrac{4 \pi \mathrm{r}^{2} \sigma}{\mathrm{r}}\right]$
$\Rightarrow \dfrac{4 \pi \sigma}{4 \pi \varepsilon_{0}}(\mathrm{R}+\mathrm{r})$
$\Rightarrow \dfrac{\sigma}{\varepsilon_{0}}(\mathrm{R}+\mathrm{r})$
Thus, the electric potential is: $\dfrac{\sigma}{\varepsilon_{0}}(\mathrm{R}+\mathrm{r})$.
Hence, the correct option is Option A.
Note: Electric potential gives us an idea about the amount of work that is required to be done in order to remove a unit of electric charge from a point to another specific point in the electric field. Thus, the total work that will be done by the external agent will give us an idea about the electric potential. The SI unit of electric potential is considered to be Volts. The main purpose of finding the electric potential is to get an idea about the source of the electric field on the basis of the location of the point that is present within the electric field. As because the electric potential is derived from the energy of the external agent it will be considered as a scalar quantity.
Complete step by step answer:
Let us draw the diagram, for a better understanding.

Let us consider that the radius of the outer sphere is R and the radius of the inner sphere is r. Thus, we can write that:
Charge on the outer sphere, $\mathrm{q}_{1}=4 \pi \mathrm{R}^{2} \sigma$ and,
Charge on the inner sphere, $\mathrm{q}_{2}=4 \pi \mathrm{r}^{2} \sigma$
Now let us consider that the electric potential is defined as v. So, the electric potential of the common centre will be given as the summation of the charges on both the outer and the inner spheres.
Hence, we can write that:
$\Rightarrow \mathrm{v}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\mathrm{q}_{1}}{\mathrm{R}}+\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\mathrm{q}_{2}}{\mathrm{r}}$
After the summation we get the value of v as:
$\Rightarrow \dfrac{1}{4 \pi \varepsilon_{0}}\left[\dfrac{4 \pi \mathrm{R}^{2} \sigma}{\mathrm{R}}+\dfrac{4 \pi \mathrm{r}^{2} \sigma}{\mathrm{r}}\right]$
$\Rightarrow \dfrac{4 \pi \sigma}{4 \pi \varepsilon_{0}}(\mathrm{R}+\mathrm{r})$
$\Rightarrow \dfrac{\sigma}{\varepsilon_{0}}(\mathrm{R}+\mathrm{r})$
Thus, the electric potential is: $\dfrac{\sigma}{\varepsilon_{0}}(\mathrm{R}+\mathrm{r})$.
Hence, the correct option is Option A.
Note: Electric potential gives us an idea about the amount of work that is required to be done in order to remove a unit of electric charge from a point to another specific point in the electric field. Thus, the total work that will be done by the external agent will give us an idea about the electric potential. The SI unit of electric potential is considered to be Volts. The main purpose of finding the electric potential is to get an idea about the source of the electric field on the basis of the location of the point that is present within the electric field. As because the electric potential is derived from the energy of the external agent it will be considered as a scalar quantity.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

