
A charged spherical conductor of radius a and charge q, is surrounded by another charged concentric sphere of radius b$(b > a)$. The potential difference between conductors is V. When, the spherical conductor of radius b is discharged completely, then the potential difference between conductor will be:
(A) V
(B) $\dfrac{{{V_a}}}{b}$
(C) $\dfrac{{{q_1}}}{{4\pi {\varepsilon _0}a}} - \dfrac{{{q_2}}}{{4\pi {\varepsilon _0}b}}$
(D) None of the these
Answer
126k+ views
Hint: we know that a spherical conductor cannot generate an electric field inside its surface by its own if an electric field exists there must be some potential we can use these concepts here.
Complete step by step answer

From the diagram, we can see that two concentric spheres, bigger one with radius b and smaller one with radius a and charge q are connected. It is given that the potential difference between them is V.
Let us assume on outer sphere the charge is ${q_1}$ at time t
Then after discharge ${q_1} = 0$
If there is potential difference, then there must be electric field in the sphere
This electric field cannot be due to outer sphere, it can be due to the inner sphere of radius ‘a’
Since the charge of the inner sphere remains ‘q’ throughout, hence the electric field between the two concentric spheres also remains the same.
Since the electric field is the same, this means that the potential difference will also remain the same that is V.
Hence, the correct option is A.
Additional information
To determine electric field on a body because of symmetric charge distribution Gauss theorem can be used. This law relates the electric field at a point on a closed surface to the enclosed charge by the surface. Electric potential is the work done on bringing a point charge from infinity that is your reference point to that point.
Note:
Electric field outside and inside the sphere is radially distributed. You should also note that the electric potential decreases in the electric field direction and is negative due to negative charge and positive due to positive charge except at infinity. The potential at infinity is zero.
Complete step by step answer

From the diagram, we can see that two concentric spheres, bigger one with radius b and smaller one with radius a and charge q are connected. It is given that the potential difference between them is V.
Let us assume on outer sphere the charge is ${q_1}$ at time t
Then after discharge ${q_1} = 0$
If there is potential difference, then there must be electric field in the sphere
This electric field cannot be due to outer sphere, it can be due to the inner sphere of radius ‘a’
Since the charge of the inner sphere remains ‘q’ throughout, hence the electric field between the two concentric spheres also remains the same.
Since the electric field is the same, this means that the potential difference will also remain the same that is V.
Hence, the correct option is A.
Additional information
To determine electric field on a body because of symmetric charge distribution Gauss theorem can be used. This law relates the electric field at a point on a closed surface to the enclosed charge by the surface. Electric potential is the work done on bringing a point charge from infinity that is your reference point to that point.
Note:
Electric field outside and inside the sphere is radially distributed. You should also note that the electric potential decreases in the electric field direction and is negative due to negative charge and positive due to positive charge except at infinity. The potential at infinity is zero.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application

Young's Double Slit Experiment Step by Step Derivation

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

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

JEE Main Login 2045: Step-by-Step Instructions and Details

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

JEE Main 2025: Derivation of Equation of Trajectory in Physics
