Coulomb law is applicable only for

Inverse square law is not satisfied

Calculate the charge on big planets

Coulomb’s Law - Vector Form, Limitations, Examples, Key Points

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.

Coulomb’s Law Formula

Coulombs law 1

In Short: F ∝ q₁q₂/d²

where,

ε is absolute permittivity,
K or ε_r is the relative permittivity or specific inductive capacity
ε₀ is the permittivity of free space.
K or ε_r 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

Coulombs law 2

\(\begin{array}{l}{{\vec{F}}_{12}}=\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r_{12}^{2}}{{\hat{r}}_{12}}; \;\;{{\vec{F}}_{12}}=-{{\vec{F}}_{21}}\end{array} \)

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

\(\begin{array}{l}\left( ie\,\,\,{{{\vec{F}}}_{12}}=-{{{\vec{F}}}_{21}} \right)\end{array} \)

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

\(\begin{array}{l}\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}+{{\vec{F}}_{3}}+……\end{array} \)

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 ε_o is 8.86 × 10^-12 C²/Nm² (or) 8.86 × 10^-12 Fm^–1

Note: Coulomb force is true only for static charges.

Coulomb’s Law – Conditions for Stability

If q is slightly displaced towards A, F_A increases in magnitude while F_B 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 F_A and F_B 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,

\(\begin{array}{l}F=\frac{1}{K}\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\end{array} \)

= constant .

2. Kr² = constant or K₁r₁² = K₂r₂²

3. If the force between two charges separated by a distance ‘r₀’ in a vacuum is the 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

\(\begin{array}{l}\frac{{{q}_{1}}+{{q}_{2}}}{2}\end{array} \)

. If the charges are q₁ and –q₂, then each will have a charge equal to

\(\begin{array}{l}\frac{{{q}_{1}}-{{q}_{2}}}{2}\end{array} \)

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

\(\begin{array}{l}F = \frac{F{{\left( {{q}_{1}}+{{q}_{2}} \right)}^{2}}}{4{{q}_{1}}{{q}_{2}}}\end{array} \)

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 / √ (μ_oε_o )

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 =
\(\begin{array}{l}[\frac{q^{2}2l }{4\pi \epsilon _{o}mg}]^{\frac{1}{3}}\end{array} \)
The tension in the thread =
\(\begin{array}{l}\sqrt{f^2 + (mg)^2}\end{array} \)
If the total system is kept in space then the angle between threads is 180° and tension in a thread is given by

T =

\(\begin{array}{l}\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}^{2}}}{4{{\ell }^{2}}}\end{array} \)

A charge Q is divided into q and (Q – q). Then electrostatic force between them is maximum when

\(\begin{array}{l}\frac{q}{Q}=\frac{1}{2}\;\ (or) \;\;\;\frac{q}{\left( Q-q \right)}=1\end{array} \)

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

\(\begin{array}{l}{{\varepsilon }_{r}}\end{array} \)

\(\begin{array}{l}K=\frac{force\;between\;two\;charge\;in\;air}{force\;between\;same\;charge\;in\;the\;medium\;at\;the\;same\;distance}\end{array} \)

\(\begin{array}{l}{{\varepsilon }_{r}}\end{array} \)

\(\begin{array}{l}=\frac{F_a}{F_m}\end{array} \)

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,

\(\begin{array}{l}F\text{ }=\text{ }\frac{\text{1}}{\text{4}\pi {{\varepsilon }_{\text{o}}}}.\frac{{{\text{q}}_{\text{1}}}{{q}_{2}}}{{{d}^{2}}}\end{array} \)

since for air or vacuum,

\(\begin{array}{l}{{\varepsilon }_{r}}\end{array} \)

= K = 1

The value of

\(\begin{array}{l}\frac{\text{1}}{\text{4}\pi {{\varepsilon }_{\text{o}}}}\end{array} \)

is equal to 9 × 10⁹ Nm²/C².

⇒ Related Topics:

Application Of The Coulombs Law

To calculate the distance and force between the two charges.
The electric field can be calculated using the coulombs law

\(\begin{array}{l}E = \frac{F}{Q_{T}}\left ( \frac{N}{C} \right )\end{array} \)

Where E = Strength of the electric field

F = Electrostatic force

Q_T= 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

Coulombs law 4

F_CA =

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q\times q}{{{a}^{2}}}\end{array} \)

along AC

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

F_CB =

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q\times q}{{{a}^{2}}}\end{array} \)

along CB

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

F =

\(\begin{array}{l}\sqrt{{{F}^{2}_{C{{A}}}}+{{F}^{2}_{C{{B}}}}+2{{F}_{CA}}\times {{F}_{CB}}\cos 120{}^\text{o}}\end{array} \)

\(\begin{array}{l}\frac{{{q}^{2}}}{4\pi ε{{a}^{2}}} = \frac{9\times {{10}^{9}}\times {{(100\times {{10}^{-6}})}^{2}}}{{{4}^{2}}}\end{array} \)

= 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 r_A from A and at a distance r_B from B.

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

Coulombs law 5

FqA =

\(\begin{array}{l}\frac{1}{4\pi{{\varepsilon}_{0}}}.\frac{q}{{{r}_{A}}^{2}}\end{array} \)

towards A

Force on q due to B

FqB=

\(\begin{array}{l}\frac{1}{4\pi{{\varepsilon}_{0}}}. \frac{q}{{{r}_{B}}^{2}}\end{array} \)

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.

|F_qA| = |F_qB|

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}} = \frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{B}}^{2}}\end{array} \)

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 =

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}}\end{array} \)

towards q

Force on A by B =

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{(1)(1)}{{{({{r}_{A}}+{{r}_{B}})}^{2}}}\end{array} \)

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{1}{{{(2{{r}_{A}})}^{2}}}\end{array} \)

away from q

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

\(\begin{array}{l}\frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{q}{{{r}_{A}}^{2}}= \frac{1}{4\pi {{\varepsilon }_{0}}}. \frac{1}{{{(2{{r}_{A}})}^{2}}}\end{array} \)

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

Given

q_{1 =} 6×10^-6C

q₂= 4×10^-6C

r = 0.040 m

Sol.

\(\begin{array}{l}F_{e} = k \frac{q_{1}q_{2}}{r^{2}}\end{array} \)

\(\begin{array}{l}F_{e} = \frac{8.99\times 10^{9}\left ( 6\times 10^{-6} \right )\left (4\times 10^{-6} \right )}{\left ( 0.04^{2} \right )}\end{array} \)

\(\begin{array}{l}F_{e} = \frac{8.99\times 10^{9}\left ( 2.4\times 10^{-11} \right )}{1.6\times 10^{-3}}\end{array} \)

\(\begin{array}{l}F_{e} = \frac{0.21576}{1.6\times 10^{-3}}\end{array} \)

F_e= 134.85 N

Problem 4: Two-point charges, q₁= +9 μC and q₂= 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

q₁= 9 ×10^-6 C

q₂ = 4 ×10^-6 C

r = 12cm = 0.12 m

Sol:

\(\begin{array}{l}F_{e} = k \frac{q_{1}q_{2}}{r^{2}}\end{array} \)

\(\begin{array}{l}F_{e} = \frac{8.99\times 10^{9}\left ( 9\times 10^{-6} \right )\left (4\times 10^{-6} \right )}{\left ( 0.12^{2} \right )}\end{array} \)

\(\begin{array}{l}F_{e} = \frac{8.99\times 10^{9}\left ( 3.6\times 10^{-11} \right )}{0.0144}\end{array} \)

\(\begin{array}{l}F_{e} = \frac{0.32364}{0.0144}\end{array} \)

F_e= 22.475 N