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
301.2k+ 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
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

What Are Elastic Collisions in One Dimension?

Effective Nuclear Charge for JEE

