Forces Between Multiple Charges

Understanding the forces between multiple charges is essential in physics, especially in the study of electrostatics. When more than two electric charges are present, each charge experiences a force due to all other charges in the system. These forces are vector quantities, meaning they have both magnitude and direction. By applying Coulomb’s Law and the principle of superposition, we can calculate the net electrostatic force acting on any individual charge. This concept is fundamental in fields such as electronics, electromagnetism, and atomic physics.

1.0Coulombs Law

• The electrostatic force of interaction between two point charges is directly proportional to the product of magnitude of charges and inversely proportional to the square of distance between them.

• This force always works along the line joining the two charges.

$\left|{\vec{F}}_{12}\right|=\left|{\vec{F}}_{21}\right|=F$

$F\propto \frac{|q_1{q}_2|}{r^2}\Rightarrow F=\frac{k|q_1{q}_2|}{r^2}$

$k=\text{ proportionality constant or coulomb’s constant or electrostatic constant }$

• The value of k depends upon the choice of system of units.
• In SI system, $k=9\times 10^9\frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}$
• This constant is also written as, $k=\frac{1}{4\pi {ϵ}_0}$

${ϵ}_0=\text{ permittivity of free space or vacuum }=8.85\times 10^{-12}\frac{C^2}{N-m^2}$

$F=\frac{k{q}_1{q}_2}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$

• When the point charge system is placed in a homogeneous dielectric medium.

$F_m=\frac{1}{4\pi ϵ}\frac{q_1{q}_2}{r^2}$

$ϵ=\text{ permittivity of medium }={ϵ}_0{ϵ}_r$

${ϵ}_r=\text{ relative permittivity of medium or Dielectric constant of medium }(K)$

$F_m=\frac{1}{4\pi {ϵ}_0{ϵ}_r}\frac{q_1{q}_2}{r^2}=\frac{1}{{ϵ}_r}\left(\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}\right)\Rightarrow F_m=\frac{F}{{ϵ}_r}$

The value of ${ϵ}_{r}orK\ge 1\text{ i.e. }{F}_m\le F$

2.0Coulomb's Law in Vector Form

${\vec{F}}_{12}=\text{ Force on }{q}_1\text{ due to }{q}_2=\frac{k{q}_1{q}_2}{r^2}\widehat{r_{21}}=\frac{k{q}_1{q}_2}{r^3}{\vec{r}}_{21}$

${\vec{F}}_{21}=\text{ Force on }{q}_2\text{ due to }{q}_1=\frac{k{q}_1{q}_2}{r^2}\widehat{r_{12}}=\frac{k{q}_1{q}_2}{r^3}{\vec{r}}_{12}$

$\because r=|\overrightarrow{AB}|=|\overrightarrow{BA}|\Rightarrow r=\left|{\vec{r}}_2-{\vec{r}}_1\right|=\left|{\vec{r}}_1-{\vec{r}}_2\right|$

${\vec{F}}_{12}=\frac{k{q}_1{q}_2}{{|{\vec{r}}_1-{\vec{r}}_2|}^3}\left({\vec{r}}_1-{\vec{r}}_2\right)\text{ and }$

${\vec{F}}_{21}=\frac{k{q}_1{q}_2}{{|{\vec{r}}_2-{\vec{r}}_1|}^3}\left({\vec{r}}_2-{\vec{r}}_1\right)$

3.0Principle of Superposition

• The superposition principle states that when multiple electric charges interact, the total force acting on a particular charge is the vector sum of the individual forces exerted on it by each of the other charges. Importantly, the force between any two charges remains unaffected by the presence of additional charges in the system.

• When a number of charges are interacting then, the total force on a given charge is the vector sum of all the individual forces exerted on it by all other charges.
  

${\vec{F}}_1={\vec{F}}_{12}+{\vec{F}}_{13}+\dots \dots \dots {\vec{F}}_{1n}$

$\vec{F}=\frac{k{q}_0{q}_1}{r_1^2}{\hat{r}}_1+\frac{k{q}_0{q}_2}{r_2^2}{\hat{r}}_2+\dots \dots +\frac{k{q}_0{q}_i}{r_i^2}{\hat{r}}_i+\dots \dots \dots \dots \frac{k{q}_0{q}_n}{r_n^2}{\hat{r}}_n$

4.0Forces Between Multiple Charges

• N point charges $q_1,q_2,q_3\dots \dots \dots q_N$ placed in vacuum,charge situated at specific location represented by a position vector from the origin O represented as $r_1,r_2,r_3\dots \dots \dots r_N$
• Total Forces on charge $q_1$is the vector sum of the forces from the other charges.

$\overrightarrow{F_1}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\dots \dots \dots \dots +\overrightarrow{F_{1N}}$

${\vec{F}}_1=\frac{k{q}_1{q}_2}{{|{\vec{r}}_1-{\vec{r}}_2|}^3}\left({\vec{r}}_1-{\vec{r}}_2\right)+\frac{k{q}_1{q}_3}{{|{\vec{r}}_1-{\vec{r}}_3|}^3}\left({\vec{r}}_1-{\vec{r}}_3\right)+\dots \dots \dots \dots +\frac{k{q}_1{q}_N}{{|{\vec{r}}_1-{\vec{r}}_N|}^3}\left({\vec{r}}_1-{\vec{r}}_N\right)$

${\vec{F}}_1=\frac{q_1}{4\pi {ϵ}_0}{\sum }_{i=2}^N{q}_i\frac{{\vec{r}}_1-{\vec{r}}_i}{{|{\vec{r}}_1-{\vec{r}}_i|}^3}$

5.0Steps to Solve Multiple Charges Problem

1. Draw a Clear Diagram: Mark all charges with correct positions, magnitudes, and signs.

2. Choose the Target Charge: Decide which charge you want to find the net force on.

3. Apply Coulomb’s Law to Each Pair: Find the magnitude of the force between the target charge and each other charge.

$F=\frac{k|q_1{q}_2|}{r^2}$

4.Determine the Direction of Each Force: Like charges repel, unlike charges attract. Indicate direction with arrows in your diagram.

5.Resolve Forces into Components :For 2D/3D problems, break forces into x- and y-components using trigonometry.

6.Add All Forces (Superposition Principle) :Add all x-components and y-components separately:

$F_x=\sum F_{xi}\text{ and }{F}_y=\sum F_{yi}$

7.Calculate the Net Force by using vector addition

$F_{net}=\sqrt{F_x^2+F_y^2}$

$\theta ={\mathrm{Tan}}^{-1}\left(\frac{F_y}{F_x}\right)$

8.Check Units and Direction:Ensure force is in Newtons (N) and direction is correct.

6.0Solved Examples on Forces Between Multiple Charges

Q-1. Three equal charges, each of magnitude +Q, are placed at the corners of an equilateral triangle of side length a. What is the net force acting on any one of the charges?

Solution: Suppose net force is to be calculated on the charge which is kept at A. Two charges kept at B and C are applying force on that particular charge, with direction as shown in the figure.

$F_B=F_C=F=k\frac{Q^2}{a^2}$

$F_{net}=\sqrt{F_B^2+F_C^2+2F_B{F}_C\cos 60^{\circ }}$

$F_{net}=\sqrt{3}F=\frac{\sqrt{3}k{Q}^2}{a^2}$

Q-2. Equal charges of magnitude Q are placed at the four corners A, B, C, and D of a square with side length a. What is the magnitude of the net force acting on the charge located at corner B?

Solution:

$F_{net}=F_{AC}+F_D=\sqrt{F_A^2+F_C^2}+F_D$

$F_A=F_C=\frac{k{Q}^2}{a^2}and{F}_D=\frac{k{Q}^2}{(a\sqrt{2}{)}^2}$

$F_{net}=\frac{\sqrt{2}k{Q}^2}{a^2}+\frac{k{Q}^2}{2a^2}=\frac{k{Q}^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)=\frac{Q^2}{4\pi {ϵ}_0{a}^2}\left(\frac{1+2\sqrt{2}}{2}\right)$

Q-3. ABC is a right-angled triangle with AB = 3 cm, BC = 4 cm, and $\angle ABC=90^{\circ }$. Three point charges of +15 e.s.u., +12 e.s.u., and –20 e.s.u. are placed at points A, B, and C respectively. What is the net force acting on the charge at point B?

Solution:

Net force on B, $F_{net}=\sqrt{F_A^2+F_C^2}$

$F_A=\frac{15\times 12}{(3{)}^2}=20\text{ dyne }$

$F_C=\frac{12\times 20}{(4{)}^2}=15\text{ dyne }$

$F_{net}=25\text{ dyne }$

Q-4. Five point charges, each of value q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value -q coulomb placed at the centre of the hexagon?

Solution:

In the diagram, we can see that force due to charge A and D are opposite to each

Other

${\vec{F}}_{OF}+{\vec{F}}_{OC}=\overrightarrow{0}......(1)$

${\vec{F}}_{OB}+{\vec{F}}_{OE}=\vec{0}........(2)$

${\vec{F}}_{net}={\vec{F}}_{OF}+{\vec{F}}_{OB}+{\vec{F}}_{OC}+{\vec{F}}_{OE}$

Using equation (1) and (2)

${\vec{F}}_{net}={\vec{F}}_{OD}=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{L^2}\text{ along }OD$

Q-5. Find the net force on the charge at B.

Solution:

So, the net force on B will be,

$F_r=\sqrt{2}\frac{k{q}^2}{r^2}$