COULOMB’S LAW

COULOMB’S LAW:

Two point electric charges q1 and q2 at rest, separated by a distance r exert a force on each other whose magnitude is given by

If between the two charges there is free space then

Where ε0 is the absolute electric permittivity of the free space and ε0 = 8.85 x 10-12 C2 / N-m2

Regarding Coulomb’s law, the following points are worth noting:

  1. Coulomb’s law stated above describes the interaction of two point charges. When two charges
    exert forces simultaneously on a third charge, the total force acting on that charge is the vector
    sum of the forces that the two charges would exert individually. This important property, called
    the principle of superposition of forces, holds for any number of charges. Thus,
    Fnet = F1 + F2 + F3 + Fn
  2. The electric force is an action reaction pair, i.e. the two charges exert equal and opposite forces on
    each other.
  3. The electric force is conservative in nature.

Illustration 1. A polythene piece rubbed with wool is found to have negative charge of 3.2 x 10-7 C.
(a) Estimate the number of electrons transferred from wool to polythene.
(b) Is there a transfer of mass from wool to polythene? If yes, how much?

Solution: (a) Let n be the number of electrons getting transferred.
⇒ n x e = 3.2 x 10-7
⇒ 1.6 x 10-19 n = 3.2 x 10-7
⇒ n = 2 x 1012 ⇒ 2 x 1012 electrons will get transferred.
(b) Mass transferred will be product of number of electrons and the mass of electron.
If m = mass getting transferred.
⇒ m = n x (9.1 x 10-31) kg
m = 2 x 1012 x 9.1 x 10-31
m = 18.2 x 10-31 kg


Illustration 2. Calculate force between two charges of 2 C each separated by 2 m in vacuum.

Solution: F = kQ1Q2/R2 = 9 x 109 x 2 x 2 ⇒ F = 9 x 109 N


Illustration 3. Two particles A and B having charges 8 x 10-6 C and –2 x10-6 C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?

Solution: As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B. Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q



Illustration 4. A charge of -2 μC is placed at the perpendicular bisector of the line joining two point charges of 10 μC as shown in figure. What is the net force acting on the -2 μC charge ?

Components of forces parallel to AB will cancel out.
F = F1 cos θ + F2 cos θ


Questions For You:

(i) A negatively charged particle is placed exactly midway between two fixed particles having equal positive charge. What will happen to the charge ?
(a) if it is displaced at right angle to the line joining the positive charges?
(b) if it is displaced along the line joining the positive charges?

(ii) Is the Coulomb force between two given charges affected in anyway, if other charges are brought in the neighbourhood ?

Coulombs law in Vector Relations


PRINCIPLE OF SUPERPOSITION
This principle tells us that if charge Q is placed in the vicinity of several charges q1, q2 ….. qn, then the force on Q can be found out by calculating separately the forces F1 ,F2 …Fn , exerted by q1, q2, ….. qn respectively on Q and then adding these forces vectorially. Their resultant F is the total force on Q due to all of charges.


Illustration 5. It is required to hold equal charges q each in equilibrium at the corners of a square of
side a. What charge when placed at the centre of the square will do this?

Solution:

Let the charge be Q
As ABCD is a square of side a, r=a/√2

Net force on the charge at B is

For charge q to be in equilibrium at B, the net force on it must be zero. Taking x-component


Questions Based on Coulomb’s Law

Q.1. Find the dimensional formula of ε0.

Answer M − 1 L − 3 T 4 A 2

Q.2. A charge of 1.0 C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2⋅0 km. Find the force exerted by the charges on each other. How many times of your weight is this force ?

Answer 2.25 × 10 3 N

Q.3. At what separation should two equal charges, 1⋅0 C each, be placed so that the force between them equals the weight of a 50 kg person ?

Answer 4.3 × 10 3 m

Q.4. Two equal charges are placed at a separation of 1⋅0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person ?

Answer 2.3 × 10 -4 C

Q.5. Find the electric force between two protons separated by a distance of 1 fermi (1 fermi = 10 − 15 m). The protons in a nucleus remain at a separation of this order.

Answer 230 N

Q.6. Is there any lower limit to the electric force between two particles placed at a separation of 1 cm ?

Answer Yes, there’s a lower limit to the electric force between two particles placed at a separation of 1 cm, which is equal to the magnitude of force of repulsion between two electrons placed at a separation of 1 cm.

Q.7. Consider two particles A and B having equal charges and placed at some distance. The particle A is slightly displaced towards B. Does the force on B increase as soon as the particle A is displaced ? Does the force on the particle A increase as soon as it is displaced ?

Answer Electrostatic force follows the inverse square law, F = K/r2 . This means that the force on two particles carrying charges increases on decreasing the distance between them. Therefore, as particle A is slightly displaced towards B, the force on B as well as A will increase.

Q.8. Does the force on a charge due to another charge depend on the charges present nearby ?

Answer No

Q.9. Suppose all the electrons of 100 g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positively charged particle. If these two particles are placed 10⋅0 cm away from each other, find the force of attraction between them. Compare it with your weight.

Answer 2.56 × 10 25 N

Q.10. Consider a gold nucleus to be a sphere of radius 6⋅9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion ?

Answer 1.2 N

Q.11. Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0⋅1 N, how many electrons were transferred from one sphere to the other during rubbing ?

Answer 2 × 10 11

Q.12. NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2⋅75 × 10 − 8 cm, find the force of attraction between them. State the assumptions (if any) that you have made.

Answer 3.05 × 10 − 9 N

Q.13. Find the ratio of the electric and gravitational forces between two protons.

Answer 1.23 × 10 36

Q.14. Ten positively charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, …, 100 cm. The first particle has a charge 1⋅0 × 10 − 8 C, the second 8 × 10 − 8 C, the third 27 × 10 − 8 C and so on. The tenth particle has a charge 1000 × 10 − 8 C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Answer 4.95 × 10 5 N

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