Screen Readable NCERT Class 12 Physics for blind and visually impaired students.

Coulomb’s law is a quantitative statement about the force between two point charges. When the linear size of charged bodies are much smaller than the distance separating them, the size may be ignored and the charged bodies are treated as point charges. Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges, and acted along the line joining the two charges. Thus, if two point charges q₁, q₂ are separated by a distance of r in vacuum, the magnitude of the force (F vector) between them is given by

\[F\ =\ k\ ×\ \frac{(q_1\ q_2)}{r^2}\ …\ …\ (1.1)\]

How did Coulomb arrive at this law from his experiments? Coulomb used a torsion balance*

(* A torsion balance is a sensitive device to measure force. It was also used later by Cavendish to measure the very feeble gravitational force between two objects, to verify Newton’s Law of Gravitation.) for measuring the force between two charged metallic spheres. When the separation between two spheres is much larger than the radius of each sphere, the charged spheres may be regarded as point charges. However, the charges on the spheres were unknown, to begin with. How then could he discover a relation like Equation (1.1)? Coulomb thought of the following simple way: Suppose the charge on a metallic sphere is q. If the sphere is put in contact with an identical uncharged sphere, the charge will spread over the two spheres. By symmetry, the charge on each sphere will be q/2*.

(* Implicit in this is the assumption of additivity of charges and conservation: two charges (q/2 each) add up to make a total charge q.)

Repeating this process, we can get charges q/2, q/4, etc. Coulomb varied the distance for a fixed pair of charges and measured the force for different separations. He then varied the charges in pairs, keeping the distance fixed for each pair. Comparing forces for different pairs of charges at different distances, Coulomb arrived at the relation, Equation (1.1).

Coulomb’s law, a simple mathematical statement, was initially experimentally arrived at in the manner described above. While the original experiments established it at a macroscopic scale, it has also been established down to subatomic level (r ~ 10^−10 m).

Coulomb discovered his law without knowing the explicit magnitude of the charge. In fact, it is the other way round: Coulomb’s law can now be employed to furnish a definition for a unit of charge. In the relation, Equation (1.1), k is so far arbitrary. We can choose any positive value of k. The choice of k determines the size of the unit of charge. In SI units, the value of k is about 9 × 10^9 (Nm^2)/C^2. The unit of charge that results from this choice is called a coulomb which we defined earlier in Section 1.4. Putting this value of k in Equation (1.1), we see that for q1 = q2 = 1 C, r = 1 m

F = 9 × 10^9 N

That is, 1 C is the charge that when placed at a distance of 1 m from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude 9 × 10^9 N. One coulomb is evidently too big a unit to be used. In practice, in electrostatics, one uses smaller units like 1 mC or 1 μC.

The constant k in Equation (1.1) is usually put as

\[k\ =\ \frac{1}{4πε_0}\]

for later convenience, so that Coulomb’s law is written as

\[F\ =\ \frac{1}{4πε_0}\ ×\ \frac{(q_1\ q_2)}{r^2}\ …\ …\ (1.2)\]

ε₀ is called the permittivity of free space. The value of ε₀ in SI units is

\[ε_0\ =\ 8.854\ ×\ 10^{−12}\ C^2\ N^{−1}\ m^{−2}\]

START OF BLUE BOX

Charles Augustin de Coulomb (1736 - 1806)

Coulomb, a French physicist, began his career as a military engineer in the West Indies. In 1776, he returned to Paris and retired to a small estate to do his scientific research. He invented a torsion balance to measure the quantity of a force and used it for determination of forces of electric attraction or repulsion between small charged spheres. He thus arrived in 1785 at the inverse square law relation, now known as Coulomb’s law. The law had been anticipated by Priestley and also by Cavendish earlier, though Cavendish never published his results. Coulomb also found the inverse square law of force between unlike and like magnetic poles.

END OF BLUE BOX

Since force is a vector, it is better to write Coulomb’s law in the vector notation. Let the position vectors of charges q_1 and q_2 be r_1 vector and r_2 vector respectively [see Fig. 1.6(a)].

FIGURE 1.6 (a) Geometry

We denote force on q₁ due to q₂ by F₁₂ vector and force on q_2 due to q_1 by F₂₁ vector. The two point charges q1 and q2 have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by r₂₁ vector:

\[\vec{r_{21}}\ =\ \vec{r_2}\ −\ \vec{r_1}\]

In the same way, the vector leading from 2 to 1 is denoted by \(\vec{r_12}

\[\vec{r_{12}}\ =\ \vec{r_1}\ −\ \vec{r_2} =\ −\vec{r_{21}}\]

The magnitude of the vectors r21 vector and r12 vector is denoted by r21 and r12, respectively (r12 = r21). The direction of a vector is specified by a unit vector along the vector. To denote the direction from 1 to 2 (or from 2 to 1), we define the unit vectors:

r21 cap = r21 vector /r21,
r12 cap = r12 vector /r12,
r21 cap = −r12 cap

Coulomb’s force law between two point charges q1 and q2 located at r1 vector and r2 vector is then expressed as

\[\vec{F_{21}}\ =\ \frac{1}{4πε_0}}\ ×\ \frac{q_1\ q_2}{r_{21}^2}\ ×\ r_{21}^{cap}\ …\ …\ (1.3)\]

Some remarks on Equation (1.3) are relevant:

• Equation (1.3) is valid for any sign of q1 and q2 whether positive or negative.

If q1 and q2 are of the same sign (either both positive or both negative), F_21 vector is along r_21 cap, which denotes repulsion, as it should be for like charges.

If q1 and q2 are of opposite signs, F_21 vector is along − r_21 cap (= r_12 cap), which denotes attraction, as expected for unlike charges.

Thus, we do not have to write separate equations for the cases of like and unlike charges. Equation (1.3) takes care of both cases correctly [Fig. 1.6(b)].

FIGURE 1.6 (b) Forces between charges.

• The force F_12 vector on charge q_1 due to charge q_2, is obtained from Equation (1.3), by simply interchanging 1 and 2, i.e.,

\[\bec{F_{12}\ =\ \frac{1}{{4πε_0}\ ×\ \frac{q_1\ q_2}{r_{12}^2}\ ×\ r_{12}^{cap}\ =\ −\vec{F_{21}}\]

Thus, Coulomb’s law agrees with the Newton’s third law of motion.

• Coulomb’s law [Equation (1.3)] gives the force between two charges q_1 and q_2 in vacuum. If the charges are placed in matter or the intervening space has matter, the situation gets complicated due to the presence of charged constituents of matter. We shall consider electrostatics in matter in the next chapter.

Example 1.4

Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.
(a) Compare the strength of these forces by determining the ratio of their magnitudes
(i) for an electron and a proton and
(ii) for two protons.

(b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (= 10^−10 m) apart?
\[m_p\ =\ 1.67\ ×\ 10^{−27}\ kg,\]
\[m_e\ =\ 9.11\ ×\ 10^{−31}\ kg\]

Solution:
(a) (i) The electric force between an electron and a proton at a distance r apart is:
F e = −1/(4πε0) × e^2/r^2
where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is:
FG = −G × (mp m e)/r^2
where m p and m e are the masses of a proton and an electron respectively.
|F e/F G| = e^2/4πε0G m p m e = 2.4 × 10^39
Where | | represents the magnitude of the vector enclosed in the | |.

(ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is:
|F e/FG| = e^2/4πε0G mp mp = 1.3 × 10^36

However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of these forces between two protons inside a nucleus (distance between two protons is ~ 10^−15 m inside a nucleus) are F e ~ 230 N, whereas, FG ~ 1.9 × 10^−34 N.
The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

(b) The electric force F vector exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however, the masses of an electron and a proton are different. Thus, the magnitude of force is
|F vector | = 1/4πε0 × e^2/r^2 = 8.987 × 10^9 Nm^2/C^2 × (1.6 ×10^−19C)^2 / (10^−10m)^2
= 2.3 × 10^−8 N

Using Newton’s second law of motion, F = m a, the acceleration that an electron will undergo is
a = 2.3 × 10^−8 N / 9.11 × 10^−31 kg
= 2.5 × 10^22 m/s^2

Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton.
The value for acceleration of the proton is

2.3 × 10^−8 N / 1.67 × 10^−27 kg
= 1.4 × 10^19 m/s^2

Example 1.5

A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.7(a)

Fig. 1.7(a)

The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.7(b).

Fig. 1.7(b)

C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.7(c).

Fig. 1.7(c)

What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.

Solution: Let the original charge on sphere A be q and that on B be q′. At a distance r between their centres, the magnitude of the electrostatic force on each is given by
F = 1/4πε0 × (q q′)/r^2
neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q ′ /2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
F′ = 1/4πε0 × [(q/2)(q′/2)]/(r/2)^2
= 1/4πε0 × (q q′)/r^2 = F
Thus the electrostatic force on A, due to B, remains unaltered.