COULOMB’S LAW

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 $q_1$, $q_2$ are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by

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 Eq. (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, Eq. (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, Eq. (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$^9 \ \frac{N.m^2}{C^2}$. 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 Eq. (1.1), we see that for $q_1 = q_2$ = 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 Eq. (1.1) is usually put as $k = \frac{1}{4πε_0}$ for later convenience, so that Coulomb’s law is written as

$F=\frac{1}{4πε_0}\frac{|q_1q_2|}{r^2}$

$ε_0$ is called the permittivity of free space . The value of ε 0 in SI units is 

$ε_0$ = 8.854 × 10$^{–12} \ C^2N^{–1}m^{–2}$

Fig 1: Geometry between charges.

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 $r_1$ and $r_2$ respectively [see Fig.1.6(a)]. We denote force on $q_1$ due to $q_2$ by $F_{12}$ and force on $q_2$ due to $q_1$ by $F_{21}$. The two point charges $q_1$ and $q_2$ have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by $r_{21}$:

Fig 2: Forces between charges..

$r_{21} = r_2 – r_1$

In the same way, the vector leading from 2 to 1 is denoted by $r_{12}$:

$r_{12} = r_1 – r_2 = – r_{21}$

The magnitude of the vectors $r_{21}$ and $r_{12}$ is denoted by $r_{21}$ and $r_{12}$, respectively ($r_{12} = r_{21}$). 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:

$\hat{\mathbf{r}}_{21}=\frac{\mathbf{r}_{21}}{r_{21}}$; $\hat{\mathbf{r}}_{12}=\frac{\mathbf{r}_{12}}{r_{12}}$, $\hat{\mathbf{r}}_{21} =- \hat{\mathbf{r}}_{12}$

Coulomb’s force law between two point charges $q_1$ and $q_2$ located at $r_1$ and $r_2$, respectively is then expressed as

$\mathbf{F}_{21}=\frac{1}{4πε_0}\frac{q_1q_2}{r^2_{21}}\hat{\mathbf{r}}_{21}$

Remarks on Equation (1.3):

• Applicability:
  Equation (1.3) is valid for any sign of $q_1$ and $q_2$, whether positive or negative.

  • If $q_1$ and $q_2$ have the same sign (both positive or both negative), the force $\mathbf{F}_{21}$ is directed along $\hat{\mathbf{r}}_{21}$, indicating repulsion, as expected for like charges.
  • If $q_1$ and $q_2$ have opposite signs, $\mathbf{F}_{21}$ is directed along $-\hat{\mathbf{r}}_{21} (= \hat{\mathbf{r}}_{12})$, indicating attraction, as expected for unlike charges.

  This unified form of Equation (1.3) accounts for both cases (like and unlike charges) without requiring separate expressions. See [Fig. 1.6(b)] for clarification.

• Symmetry of Forces:
  The force $\mathbf{F}_{12}$ on charge $q_1$ due to $q_2$ can be derived from Equation (1.3) by interchanging the indices 1 and 2:

  $$\mathbf{F}_{12} = -\mathbf{F}_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{\mathbf{r}}_{12}$$

  This symmetry confirms that Coulomb’s law is consistent with Newton’s third law of motion.

• Force in Matter:
  Equation (1.3) describes the force between two charges $q_1$ and $q_2$ in a vacuum. When the charges are placed in a material medium, the interaction becomes more complex due to the presence of charged particles in the medium. This scenario, involving electrostatics in matter, will be explored in the next chapter.

Example 1

What is the force between two small charged spheres having charges of 2 × 10$^{–7}$ C and 3 × 10$^{–7}$ C placed 30 cm apart in air?

Solution

To calculate the electrostatic force between the two charged spheres, we use Coulomb's law:

$F=\frac{1}{4πε_0}\frac{|q_1q_2|}{r^2}$

Where:

$$q_1$ =2 × 10$^{–7}$ C

$$q_1$ =3 × 10$^{–7}$ C

$$r$ =30 cm = 0.03 m

Substituting the values:

$F=\frac{1}{4πε_0}\frac{(2 \times 10^{-7} \ C)(3 \times 10^{-7} \ C)}{(0.03)^2}$

$F=6.003 \times 10^{-3}$ N = 6.0 mN

Example 2

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:

$F_e=-\frac{1}{4πε_0}\frac{e^2}{r^2}$

where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is:

$F_G=-\frac{1}{4πε_0}\frac{m_p.m_e}{r^2}$

where $m_p$ and $m_e$ are the masses of a proton and an electron respectively.

$\left| \frac{F_e}{F_G}\right|=\frac{e^2}{4πε_0 m_pm_e}=2.4 \times 10^{39}$

(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:

$\left| \frac{F_e}{F_G}\right|=\frac{e^2}{4πε_0 m_pm_p}=1.3 \times 10^{36}$

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 Fe ~ 230 N, whereas, $F_G$ ~ 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 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 

$|\mathbf{F}|=\frac{1}{4πε_0}\frac{e^2}{r^2}$

$|\mathbf{F}|=(8.987 \times 10^9 \ Nm^2/C^2)\frac{(1.6 \times 10^{-19} \ C)^2}{(10^{-10 \ m})^2}$

$=2.3 \times 10^{-8} \ N$ 

Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is

$a_e=\frac{F}{m}=\frac{2.3 \times 10^{-8}}{9.11 \times 10^{-31}}=2.5 \times 10^{22} \ m/s^2$

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

$a_p=\frac{F}{m}=\frac{2.3 \times 10^{-8}}{1.67 \times 10^{-27}}=1.4 \times 10^{19} \ m/s^2$

Example 3

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). 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). 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). 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

$F=\frac{1}{4πε_0}\frac{qq'}{r^2}$

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

$F'=\frac{1}{4πε_0}\frac{(q/2)(q'/2)}{(r/2)^2}=\frac{1}{4πε_0}\frac{qq'}{r^2}=F$

Thus the electrostatic force on A, due to B, remains unaltered.

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