
The electric potential at a point in space due to charge $Q$ coulomb is $Q \times {10^{11}}V$. The value of electric field due to charge $Q$ at that point is equal to:
(A) $12\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$
(B) $4\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$
(C) $12\pi { \in _0}Q \times {10^{20}}V{m^{ - 1}}$
(D) $4\pi { \in _0}Q \times {10^{20}}V{m^{ - 1}}$
Answer
232.8k+ views
Hint Electric potential at a point in space is directly proportional to charge of source charge and inversely proportional to distance between source charge and that point. An electric field at a point is directly proportional to the charge of source charge and inversely proportional to square of distance between the source charge and that point. Use this relation between electric potential and electric field to find electric field using electric potential at that point.
Complete step by step solution
Electric potential at a point in space is directly proportional to charge of source charge and inversely proportional to distance between source charge and that point.
\[V = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{R}\] (1)
Where $R$ is distance between source charge $Q$ and the given point.
an electric field at a point is directly proportional to the charge of source charge and inversely proportional to square of distance between the source charge and that point.
$E = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{{{R^2}}}$ (2)
Combining equation (1) and (2), we have
$E = \dfrac{{4\pi { \in _0}{V^2}}}{Q}$
Substituting $V = Q \times {10^{11}}$ as given in question, we get
$E = \dfrac{{4\pi { \in _0}{{(Q \times {{10}^{11}})}^2}}}{Q} = 4\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$
Hence, the correct answer is option B.
Note Electric field at a point in space due to charge $Q$ gives the value of force applied on unit positive charge placed at that point, where electric potential is potential energy of a unit positive charge at that point due to charge $Q$.
Complete step by step solution
Electric potential at a point in space is directly proportional to charge of source charge and inversely proportional to distance between source charge and that point.
\[V = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{R}\] (1)
Where $R$ is distance between source charge $Q$ and the given point.
an electric field at a point is directly proportional to the charge of source charge and inversely proportional to square of distance between the source charge and that point.
$E = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{{{R^2}}}$ (2)
Combining equation (1) and (2), we have
$E = \dfrac{{4\pi { \in _0}{V^2}}}{Q}$
Substituting $V = Q \times {10^{11}}$ as given in question, we get
$E = \dfrac{{4\pi { \in _0}{{(Q \times {{10}^{11}})}^2}}}{Q} = 4\pi { \in _0}Q \times {10^{22}}V{m^{ - 1}}$
Hence, the correct answer is option B.
Note Electric field at a point in space due to charge $Q$ gives the value of force applied on unit positive charge placed at that point, where electric potential is potential energy of a unit positive charge at that point due to charge $Q$.
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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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 Uniform Acceleration in Physics

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

