Coulomb's Law

What is Coulomb’s Law?

According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges.

Table of Content

Coulomb’s Law Formula: F ∝ q1q2/d2

Coulomb's Law Derivation

where,

  • ε is absolute permittivity,
  • K or εr is the relative permittivity or specific inductive capacity 
  • ε0 is the permittivity of free space.
  • K or εr is also called a dielectric constant of the medium in which the two charges are placed.

History of Coulomb’s Law

A French physicist Charles Augustin de Coulomb in 1785 coined a tangible relationship in mathematical form between two bodies that have been electrically charged. He published an equation for the force causing the bodies to attract or repel each other which is known as Coulomb’s law or Coulomb’s inverse-square law.

Coulomb’s Law in Vector Form

Coulomb's Law in Vector Form

 

F12=14π0q1q2r122r^12;    F12=F21{{\vec{F}}_{12}}=\frac{1}{4\pi {{\in }_{0}}}-\frac{{{q}_{1}}{{q}_{2}}}{r_{12}^{2}}{{\hat{r}}_{12}}; \;\;{{\vec{F}}_{12}}=-{{\vec{F}}_{21}}

Here F12 is the force exerted by q1 on q2 and F21 is the force exerted by q2 on q1.

Coulomb’s law holds for stationary charges only which are point sized. This law obeys Newton’s third law

(ieF12=F21)\left( ie\,\,\,{{{\vec{F}}}_{12}}=-{{{\vec{F}}}_{21}} \right)

Force on a charged particle due to a number of point charges is the resultant of forces due to individual point charges i.e.

F=F1+F2+F3+\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}+{{\vec{F}}_{3}}+……

What is 1 Coulomb of Charge?

A coulomb is that charge which repels an equal charge of the same sign with a force of 9×109 N when the charges are one meter apart in a vacuum. Coulomb force is the conservative mutual and internal force.

The value of εo is 8.86 × 10-12 C2/Nm2 (or) 8.86 × 10-12 Fm–1

Note: Coulomb force is true only for static charges.

Coulomb force

Coulomb’s Law – Conditions for Stability

If q is slightly displaced towards A, FA increases in magnitude while FB decreases in magnitude. Now the net force on q is toward A so it will not return to its original position. So for axial displacement, the equilibrium is unstable.

If q is displaced perpendicular to AB, the force FA and FB bring the charge to its original position. So for perpendicular displacement, the equilibrium is stable.

Key Points on Coulomb’s Law

1. If the force between two charges in two different media is the same for different separations, F=1K14π0q1q2r2F=\frac{1}{K}\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} = constant .

2.  Kr2 = constant or K1r12 = K2r2

3. If the force between two charges separated by a distance ‘r0’ in a vacuum is same as the force between the same charges separated by a distance ‘r’ in a medium, then from Coulomb’s Law; Kr2 = r02

4. Two identical conductors having charges q1 and q2 are put to contact and then separated after which each will have a charge equal to q1+q22\frac{{{q}_{1}}+{{q}_{2}}}{2}. If the charges are q1 and –q2, then each will have a charge equal to q1q22\frac{{{q}_{1}}-{{q}_{2}}}{2}.

5. Two spherical conductors having charges q1 and q2 and radii r1 and r2 are put to contact and then separated the charges of the conductors after contact is;

q1 = [r1/(r1 + r2)] (q1 + q2) and q2 = [r2/(r1 + r2)] (q1 + q2)

6. If the force of attraction or repulsion between two identical conductors having charges q1 and q2 when separated by a distance d is F. Also if they are put to contact and then separated by the same distance the new force between them is F=F(q1+q2)24q1q2F = \frac{F{{\left( {{q}_{1}}+{{q}_{2}} \right)}^{2}}}{4{{q}_{1}}{{q}_{2}}}

7. If charges are q1 and -q2 then, F = F(q1 + q2)2 / 4q1q2

8. Between two-electrons separated by a certain distance: Electrical force/Gravitational force = 1042

9. Between two protons separated by a certain distance: Electrical force/Gravitational force = 1036

10. Between a proton and an electron separated by a certain distance: Electrical force/Gravitational force = 1039

11. The relationship between the velocity of light, the permeability of free space and permittivity of free space is given by the expression c = 1 / √ (μoεo )

12. If Coulomb’s law is applied to two identical balls of mass m are hung by silk thread of length ‘l’ from the same hook and carry similar charges q then;

  • The distance between balls = [q22l4πϵomg]13[\frac{q^{2}2l }{4\pi \epsilon _{o}mg}]\frac{1}{3}
  • The tension in the thread = √ (f 2) + (mg)2
  • If total system is kept in space then angle between threads is 180° and tension in thread is given by T =14π0q242\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}^{2}}}{4{{\ell }^{2}}}
  • A charge Q is divided into q and (Q – q). Then electrostatic force between them is maximum when qQ=12   (or)      q(Qq)=1\frac{q}{Q}=\frac{1}{2}\;\ (or) \;\;\;\frac{q}{\left( Q-q \right)}=1

Limitations of Coulomb’s Law

  • The law is applicable only for the point charges at rest.
  • Coulomb’s Law can be only applied in those cases where the inverse square law is obeyed.
  • It is difficult to implement Coulomb’s law where charges are in arbitrary shape because in such cases we cannot determine the distance’ between the charges.
  • The law can’t be used directly to calculate the charge on the big planets.

Relative Permittivity of a Material

εr{{\varepsilon }_{r}} = K=force  between  two  charge  in  airforce  between  same  charge  in  the  medium  at  the  same  distanceK=\frac{force\;between\;two\;charge\;in\;air}{force\;between\;same\;charge\;in\;the\;medium\;at\;the\;same\;distance}.

εr{{\varepsilon }_{r}} = =FaFm=\frac{F_a}{F_m}
  • For air K = 1
  • For metals K = infinity

Force between 2 charges depends upon the nature of the intervening medium, whereas gravitational force is independent of intervening medium.

For air or vacuum, F = 14πεo.q1q2d2F\text{ }=\text{ }\frac{\text{1}}{\text{4}\pi {{\varepsilon }_{\text{o}}}}.\frac{{{\text{q}}_{\text{1}}}{{q}_{2}}}{{{d}^{2}}} since for air or vacuum, εr{{\varepsilon }_{r}} = K = 1

The value of 14πεo\frac{\text{1}}{\text{4}\pi {{\varepsilon }_{\text{o}}}} is equal to 9 × 109 Nm2/C2.

⇒ Related Topics:

Application Of The Coulombs Law

  • To claculate the distance and force between the two charges.
  • The electric field can be calculated using the coulombs law
E=FQT(NC)E = \frac{F}{Q_{T}}\left ( \frac{N}{C} \right )

Where E = Sterngth of the electric field

F = Electrostatic force

QT = Test charge in coulombs

  • To calculate the force on one point due to the presence of several points (Theorem of superposition).

Problems on Coulombs Law

Problem 1: Charges of magnitude 100 microcoulomb each are located in vacuum at the corners A, B and C of an equilateral triangle measuring 4 meters on each side. If the charge at A and C are positive and the charge B negative, what is the magnitude and direction of the total force on the charge at C?

Sol.

The situation is shown in fig. Let us consider the forces acting on C due to A and B.

Now, from Coulomb’s law, the force of repulsion on C due to A i.e., FCA in direction AC is given by

Problems on Coulombs Law 01

FCA = 14πε0.q×qa2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q\times q}{{{a}^{2}}} along AC

The force of attraction on C due to B i.e., FCB in direction CB is given by

FCB = 14πε0.q×qa2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q\times q}{{{a}^{2}}} along CB

Thus the two forces are equal in magnitude. The angle between them is 120º. The resultant force F is given by

F = FCA2+FCB2+2FCA×FCBcos120o\sqrt{{{F}_{C{{A}^{2}}}}+{{F}_{C{{B}^{2}}}}+2{{F}_{CA}}\times {{F}_{CB}}\cos 120{}^\text{o}}

= q24π6a2=9×109×(100×106)242\frac{{{q}^{2}}}{4\pi 6{{a}^{2}}} = \frac{9\times {{10}^{9}}\times {{(100\times {{10}^{-6}})}^{2}}}{{{4}^{2}}} = 5.625 Newton

This force is parallel to AB.

Problem 2: The negative point charges of unit magnitude and a positive point charge q are placed along the straight line. At what position and for what value of q will the system be in equilibrium? Check whether it is stable, unstable or neutral equilibrium.

Sol.

The two negative charges A and B of unit magnitude are shown in fig. Let the positive charge q be at a distance rA from A and at a distance rB from B.

Now, from coulombs law, Force on q due to A

Problems on Coulombs Law 02

FqA = 14πε0.qrA2\frac{1}{4\pi{{\varepsilon}_{0}}}.\frac{q}{{{r}_{A}}^{2}} towards A

Force on q due to B

FqB=14πε0.qrB2\frac{1}{4\pi{{\varepsilon}_{0}}}. \frac{q}{{{r}_{B}}^{2}} towards B.

These two forces acting on q are opposite and collinear. For the equilibrium of q, the two forces must also be equal i.e.

|FqA| = |FqB|

or 14πε0.qrA2=14πε0.qrB2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}} = \frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{B}}^{2}} Hence rA = rB

So for the equilibrium of q, it must be equidistant from A & B i.e. at the middle of AB

Now for the equilibrium of the system, A and B must be in equilibrium.

For the equilibrium of A

Force on A by q = 14πε0.qrA2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}} towards q

Force on A by B = 14πε0.(1)(1)(rA+rB)2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{(1)(1)}{{{({{r}_{A}}+{{r}_{B}})}^{2}}}

= 14πε0.1(2rA)2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{1}{{{(2{{r}_{A}})}^{2}}} away from q

The two forces are opposite and collinear. For equilibrium the forces must be equal, opposite and collinear. Hence

14πε0.qrA2=14πε0.1(2rA)2\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}}= \frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{1}{{{(2{{r}_{A}})}^{2}}}

or q = 1/4 in magnitude of either charge.

It can also be shown that for the equilibrium of B, the magnitude of q must be 1/4 of the magnitude of either charge.

 

Problem 3: A positive charge of 6×10-6 C is 0.040m from the second positive charge of 4×10-6 C. Calculate the force between the charges.

Given

q1 =  6×10-6 C

q= 4×10-6 C

r = 0.040 m

Sol.

 

Fe=kq1q2r2F_{e} = k \frac{q_{1}q_{2}}{r^{2}}

 

Fe=8.99×109(6×106)(4×106)(0.042)F_{e} = \frac{8.99\times 10^{9}\left ( 6\times 10^{-6} \right )\left (4\times 10^{-6} \right )}{\left ( 0.04^{2} \right )}

 

Fe=8.99×109(2.4×1011)1.6×103F_{e} = \frac{8.99\times 10^{9}\left ( 2.4\times 10^{-11} \right )}{1.6\times 10^{-3}}

 

Fe=0.215761.6×103F_{e} = \frac{0.21576}{1.6\times 10^{-3}}

 

F= 134.85 N

 

Problem 4: Two-point charges, q= +9 μC and q= 4 μC, are separated by a distance r = 12 cm. What is the magnitude of the electric force? 

given 

k = 8.988 x 109 Nm2C−2

q= 9 ×10-6 C

q2 = 4 ×10-6 C

r = 12cm = 0.12 m

Sol:

Fe=kq1q2r2F_{e} = k \frac{q_{1}q_{2}}{r^{2}}

 

Fe=8.99×109(9×106)(4×106)(0.122)F_{e} = \frac{8.99\times 10^{9}\left ( 9\times 10^{-6} \right )\left (4\times 10^{-6} \right )}{\left ( 0.12^{2} \right )}

 

Fe=8.99×109(3.6×1011)0.0144F_{e} = \frac{8.99\times 10^{9}\left ( 3.6\times 10^{-11} \right )}{0.0144}

 

Fe=0.323640.0144F_{e} = \frac{0.32364}{0.0144}

 

F= 22.475 N