Coulombs law – Vector Form, Limitations, Examples, Key Points

Coulomb’s law – Vector Form, Limitations, Examples, Key Points

In the year 1786, Coulomb deduced the expression for the force between two stationary point charges in vacuum or free space. Consider two point charges q1 and q2 at rest in vacuum, and separated by a distance of r as shown in Figure 1.2. According to Coulomb, the force on the point charge q2 exerted by another point charge q1 is

\(\vec{F}_{21}\) = k \(\frac{q_{1} q_{2}}{r^{2}} \hat{r}_{12}\) ………… (1.2)

where \(\hat{r}_{12}\) is the unit vector directed from charge q1 to charge q2 and k is the proportionality constant.

Coulomb’s law img 1

Coulomb’s law

Important aspects of Coulomb’s law

(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge q2 exerted by the charge q1 always lies along the line joining the two charges. \(\hat{r}_{12}\) is the unit vector pointing from charge q1 to q2. It is shown in the Figure 1.2. Likewise, the force on the charge q1 exerted by q2 is along – \(\hat{r}_{12}\) (i.e., in the direction opposite to \(\hat{r}_{12}\)).

(iii) In SI units, k = \(\frac{1}{4 \pi \epsilon_{o}}\) and its value is 9 × 109Nm2C-2. Here ∈0 is the permittivity of free space or vacuum and its value is

0 = \(\frac{1}{4πk}\) = 8.85 × 10-12C2N-1m-2.

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:

|F| = \(\frac{9 \times 10^{9} \times 1 \times 1}{1^{2}}\) = 9 × 109N.

This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of µC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec{F}_{21}\) = \(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{12}\). In a medium permitivity ∈, the force between two point charges is given by \(\vec{F}_{21}\) = \(\frac{1}{4 \pi \epsilon} \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{12}\).

Since ∈>∈0, the force between two point charges in a medium other than vaccum is always less than that in vaccum. We define the relative permitivity for a given medium as ∈r = \(\frac{\boldsymbol{\epsilon}}{\boldsymbol{\epsilon}_{0}}\). For vaccum or air, ∈r = 1 and for all other media ∈r > 1.

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses. But there are some important differences between these two laws.

The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.

The value of the gravitational constant G = 6.67 × 10-11 N m2kg-2. The value of the constant k in Coulomb law is k = 9 × 109 N m2 C-2. Since k is much more greater than G, the electrostatic force is always greater in magnitude than gravitational force for smaller size objects.

The gravitational force between two masses is independent of the medium. For example, if 1 kg of two masses are kept in air or inside water, the gravitational force between two masses remains the same. But the electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.

(vii) The force on a charge q1 exerted by a point charge q2 is given by

\(\vec{F}_{12}\) = \(\frac{1}{4 \pi \epsilon_{0}}\)\(\frac{q_{1} q_{2}}{r^{2}} \hat{r}_{21}\) Here, \(\hat{r}_{21}\) is the unit vector from charge q2 to q1.

But \(\hat{r}_{21}\) = – \(\hat{r}_{12}\),
\(\vec{F}_{12}\) = \(\frac{1}{4 \pi \epsilon_{0}}\)\(\frac{q_{1} q_{2}}{r^{2}}\left(-\hat{r}_{12}\right)\) = \(\frac{1}{4 \pi \epsilon_{0}}\)\(\frac{q_{1} q_{2}}{r^{2}}\left(\hat{r}_{12}\right)\)
(or) \(\bar{F}_{12}\) = – \(\bar{F}_{21}\)

Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

Coulomb’s law

Example 1.2

Consider two point charges q1 and q2 at rest as shown in the figure.

Coulomb’s law img 2

They are separated by a distance of 1m. Calculate the force experienced by the two charges for the following cases:

(a) q1 = +2 μC and q2 = +3 μC
(b) q1 = +2 μC and q2 = -3 μC
(c) q1= +2 μC and q2 = -3 μC kept in water (∈r = 80)

Solution:

Coulomb’s law img 3

(a) q1 = +2 μC, q2 = + 3 μC, and r = 1m.
Both are positive charges. So the force will be repulsive.
Force experienced by the charge q2 due to q1 is given by
\(\vec{F}_{21}\) = \(\frac{1}{4 \pi \epsilon_{0}}\)\(\frac{q_{1} q_{2}}{r^{2}} \hat{r}_{12}\)
Here \(\hat{r}_{12}\) is the unit vector from q1 to q2. Since q2 is located on the right of q1, we have
\(\hat{r}_{12}\) = \(\hat{i}\) and \(\frac{1}{4 \pi \epsilon_{0}}\) = 9 × 109 so that

Coulomb’s law img 4

According to Newton’s third law, the force experienced by the charge q1 due to q2 is \(\vec{F}_{12}\) = – \(\vec{F}_{21}\). Therefore
\(\vec{F}_{12}\) = – 54 × 10-3\(\hat {i} \)N.

The directions of \(\vec{F}_{21}\) and \(\vec{F}_{12}\) are shown in the above figure in case (a)

(b) q1 = +2 µC, q2 = – 3 µC, and r = 1m.
They are unlike charges. So the force will be attractive.
Force experienced by the charge q2 due to q1 is given by

Coulomb’s law img 5

The charge q2 will experience an attractive force towards q1 which is in the negative x direction.

According to Newton’s third law, the force experienced by the charge q1 due to q2 is \(\vec{F}_{12}\) = – \(\vec{F}_{21}\). Therefore,

\(\vec{F}_{12}\) = 54 × 10-3\(\hat {i} \)N

The directions of \(\vec{F}_{21}\) and \(\vec{F}_{12}\) are shown in the figure (case (b)).

(c) If these two charges are kept inside the water, then the force experienced by q2 due to q1.

Coulomb’s law img 6

Therefore,

Coulomb’s law img 7
= -0.675 × 10-3N\(\hat {i} \)

Coulomb’s law

Example 1.3

Two small-sized identical equally charged spheres, each having mass 1 g are hanging in equilibrium as shown in the figure. The length of each string is 10 cm and the angle θ is 30° with the vertical. Calculate the magnitude of the charge in each sphere. (Take g = 10 ms-2)

Coulomb’s law img 8

Solution:

If the two spheres are neutral, the angle between them will be 0° when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical.

At equilibrium, each charge experiences zero net force in each direction. We can draw a free body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions.

The free body diagram is shown below.

Coulomb’s law img 9

In the x-direction, the acceleration of the charged sphere is zero.

Using Newton’s second law (\(\vec{F}_{\text {tot }}\) = m\(\vec {a} \)), we have

Tsinθ\(\hat {i} \) – Fc\(\hat {i} \) = 0
T sinθ = Fc ………… (1)

Here T is the tension acting on the charge due to the string and Fc is the electrostatic force between the two charges. In they-direction also, the net acceleration experienced by the charge is zero.

T cos θ\(\hat {j} \) – mg \(\hat {j} \) = 0
T cos θ = mg ……………. (2)

By dividing equation (1) by equation (2),
tan θ = \(\frac{F_{e}}{m g}\)
Since they are equally charged, the magnitude of the electrostatic force is
Coulomb’s law img 10
Here r = 2a = 2Lsinθ. By substituting these values in equation (3),
Coulomb’s law img 11
Rearraging the equation (4) to get q
Coulomb’s law img 12

Coulomb’s law

Example 1.4

Calculate the electrostatic force and gravitational force between the proton and the electron in a hydrogen atom. They are separated by a distance of 5.3 × 10-11m. The magnitude of charges on the electron and proton are 1.6 × 10-19 C. Mass of the electron is me = 9.1 × 10-31 kg and mass of proton is mp = 1.6 × 10-27 kg.

Solution

The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by

Coulomb’s law img 13

The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is

Coulomb’s law img 14

The ratio of the two forces

Coulomb’s law img 15

Note that Fe ~ 1039FG

The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This is shown in Figure 1.3

Superposition Principle

Coulomb’s law explains the interaction between two point charges. If there are more than two charges, the force on one charge due to all the other charges needs to be calculated. Coulomb’s law alone does not give the answer. The superposition principle explains the interaction between multiple charges.

According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges. Consider a system of n charges, namely q1, q2, q3, ………. qn. The force on q1 and q2. The electrostatic force between two charges is not affected by the presence of other charges in the neighbourhood.

The force on q1 exerted by the charge q3 is
\(\vec{F}_{12}\) = k\(\frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}\)

where \(\hat {r} \)21 is the unit vector from q2 to q1 along the line joining the two charges and r21 is the distance between the charges q1 and q2. The electrostatic force between two charges is not affected by the presence of other charges in the neighbourhood.

The force on q1 exerted by the charge q3 is \(\vec{F}_{13}\) = k\(\frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{31}\)

By continuing this, the total force acting on the charge q1 due to all other charges is given by

Coulomb’s law img 16

Example 1.5

Consider four equal charges q1, q2, q3 and q4 = q = +1 μC located at four different points on a circle of radius 1m, as shown in the figure. Calculate the total force acting on the charge q1 due to all the other charges.

Coulomb’s law img 17

Solution

According to the superposition principle, the total electrostatic force on charge q1 is the vector sum of the forces due to the other charges,

Coulomb’s law img 18

The following diagram shows the direction of each force on the charge q1.

Coulomb’s law img 19

The charges q2 and q4 are equidistant from q1. As a result the strengths (magnitude) of the forces \(\vec{F}_{12}\) and \(\vec{F}_{14}\) are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. But the charge q3 is located farther compared to q2 and q4.

Coulomb’s law

Since the strength of the electrostatic force decreases as distance increases, the strength of the force \(\vec{F}_{13}\) is lesser than that of forces \(\vec{F}_{12}\) and \(\vec{F}_{14}\). Hence the vector representing the force \(\vec{F}_{13}\) is drawn with smaller length compared to that for forces \(\vec{F}_{12}\) and \(\vec{F}_{14}\).

From the figure, r21 = \(\sqrt{2}\)m = r41 and r21 = 2m
The magnitudes of the forces are given by
Coulomb’s law img 20

From the figure, the angle θ = 45°. In terms of the components, we have

Coulomb’s law img 21

Then the total force on q1 is

Coulomb’s law img 22

Since F12 = F14, the jth component is zero. Hence we have
\(\vec{F}_{1}^{\text {tot }}\) = (F12cos θ + F13 + F14 cos θ)\(\hat{i}\)
Substituting the values in the above equation,
Coulomb’s law img 23

The resultant force is along the positive x axis.

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