Two point charges with charges $3$ micro coulombs and $4$ micro coulombs are separated by $2\;cm$. The value of the force between them?
(A) $600\;N$
(B) $300\;N$
(C) $540\;N$
(D) $270\;N$
(E) $400\;N$
Answer
300.3k+ views
Hint: We have two point charges having $3$ micro coulomb and $4$ micro coulomb each. The distance between both the charges is given by $2\;cm$. We have to find the force between them. This question is a direct application of Coulomb’s law and can be easily solved by applying coulomb’s law. The values are all given.
Formula used:
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$
where, $F$ stands for the force between the two charges, ${\varepsilon _0}$ is the permittivity of free space, ${Q_1}$ and ${Q_2}$ are the two charges and $r$ stands for the distance between the two charges.
Complete step by step solution:
Both charges are separated by a distance.
The value of the first charge is given as, ${Q_1} = 3\mu C$
Converting into Coulomb by multiplying with ${10^{ - 6}}$, ${Q_1} = 3 \times {10^{ - 6}}C$
The value of the second charge is given as, ${Q_2} = 4\mu C$
Converting into Coulomb by multiplying with ${10^{ - 6}}$, ${Q_2} = 4 \times {10^{ - 6}}$
The distance between both charges is given as, $r = 2cm = 0.02m$
Coulomb’s law is given by,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$
Substituting the values within the above equation, we get
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\left( {3 \times {{10}^{ - 6}}} \right) \times \left( {4 \times {{10}^{ - 6}}} \right)}}{{{{\left( {0.02} \right)}^2}}}$
The value of $\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
Substituting within the above equation, we get
$F = \dfrac{{9 \times {{10}^9} \times \left( {3 \times {{10}^{ - 6}}} \right) \times \left( {4 \times {{10}^{ - 6}}} \right)}}{{\left( {0.02} \right)}} = 270N$
The answer is: Option (D): $270\;N$
Additional Information:
The magnitude of coulomb charges will depend on three factors that are the distance between the charges, the number of charges, and the nature of the media between the charges. Positive charges are attractive in nature meanwhile negative charges are repulsive in nature.
Note: Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. Like charges will have a repulsive force between them and unlike charges will have an attractive force between them.
Formula used:
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$
where, $F$ stands for the force between the two charges, ${\varepsilon _0}$ is the permittivity of free space, ${Q_1}$ and ${Q_2}$ are the two charges and $r$ stands for the distance between the two charges.
Complete step by step solution:
Both charges are separated by a distance.
The value of the first charge is given as, ${Q_1} = 3\mu C$
Converting into Coulomb by multiplying with ${10^{ - 6}}$, ${Q_1} = 3 \times {10^{ - 6}}C$
The value of the second charge is given as, ${Q_2} = 4\mu C$
Converting into Coulomb by multiplying with ${10^{ - 6}}$, ${Q_2} = 4 \times {10^{ - 6}}$
The distance between both charges is given as, $r = 2cm = 0.02m$
Coulomb’s law is given by,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$
Substituting the values within the above equation, we get
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\left( {3 \times {{10}^{ - 6}}} \right) \times \left( {4 \times {{10}^{ - 6}}} \right)}}{{{{\left( {0.02} \right)}^2}}}$
The value of $\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
Substituting within the above equation, we get
$F = \dfrac{{9 \times {{10}^9} \times \left( {3 \times {{10}^{ - 6}}} \right) \times \left( {4 \times {{10}^{ - 6}}} \right)}}{{\left( {0.02} \right)}} = 270N$
The answer is: Option (D): $270\;N$
Additional Information:
The magnitude of coulomb charges will depend on three factors that are the distance between the charges, the number of charges, and the nature of the media between the charges. Positive charges are attractive in nature meanwhile negative charges are repulsive in nature.
Note: Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. Like charges will have a repulsive force between them and unlike charges will have an attractive force between them.
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

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

Effective Nuclear Charge for JEE

Understanding the Angle of Deviation in a Prism

