VECTOR FORM OF COULOMB'S LAW
The physical quantities are of two types namely scalars (with the only magnitude) and vectors (those quantities with magnitude and direction). Force is a vector quantity as it has both magnitude and direction. The Coulomb’s law can be re-written in the form of vectors. Remember we denote vector “F” asvector r as and so on.
Let there be two charges q1 and q2, with position vectorsandrespectively. Now since both the charges are of the same sign, there will be a repulsive force between them. Let the force on the q1 charge due to q2 beand force on q2 charge due to q1 charge be The corresponding vector from q1 to q2 is
Note that by triangular law of vector addition:
Similarly, vector leading from q2 to q1 is denoted by
The direction of a vector is specified by a unit vector along the vector
Now,Coulomb's force law between two point charges q1 and q2 when force is acting on q2 due to q1 is expressed as:
Forces acting on both point charge due to each other are always equal in magnitude and opposite in direction.
Note: That, here we have assumed the charges to be kept in vacuum. If these charges are kept in a medium with permittivity ε, then the electrostatic force between q1 and q2 will be:
Case (a): If q1 and q2 are same type of charges i.e. q1 > 0, q2 > 0 or q1 < 0, q2 < 0 then the electrostatic force between q1 and q2 will be repulsive.
Case (b): If q1 and q2 are opposite types of charges i.e. q1 > 0, q2 < 0 or q1 < 0, q2 > 0 then the electrostatic force between q1 and q2 will be attractive.
Principle of Superposition
Coulomb forces follow principle of Superposition. i.e. the resultant force on anyone of the charge is equal to the vector sum of the forces exerted by the various individual charges. The individual forces are unaffected due to the presence of other charges.
For example: If four charges are present q1, q2, q3, q4, then the resultant force exerted by charge 2, 3 and 4 on charge 1 is
Where is the net electrostatic force acting on charge q1is the electrostatic force on q1 due to charge q2, is the electrostatic force on q1 due to charge q3 and is the electrostatic force on q1 due to charge q4.
Q.1. Two point charges of magnitude 1 C and – 3 C are kept at a distance of 3 m. Find the force of attraction between them.
Ans. We have q1 = 1C, q2 = – 3C and r = 3m. Then using Coulomb’s Law and substituting above values we get
Substituting the given values,
F = 3 × 109 N
Q.2. The force between two identical charges separated by 1 cm is equal to 90 N. What is the magnitude of the two charges?
Ans. First, draw a force diagram of the problem.
Fig: Two charges separated by one centimeter experiencing a force of repulsion of 90 N
Given, F = 90 N
q1 = charge of first body = q2 = charge of second body = q
r = 1 cm
Use the Coulomb’s Law equation
Substituting the values,
Since we want the magnitude of charges, solve for q
q = ± 1.00 × 10-6 C
Since the force is repulsive and the charges are identical, they can be either both positive or both negative.
Two identical charges of ± 1.00 × 10-6 C separated by 1 cm produce a repulsive force of 90 N.
Q.3. What is the ratio of electrostatic force to the gravitational force between an electron and a proton kept at a distance r from each other? Given: mₑ=9.11×10⁻³¹ kg, mₚ=1.67×10⁻²⁷ kg, k=9×10⁹ Nm²/C², G=6.67×10⁻¹¹ Nm²/kg²
Ans. We can treat the electron and the proton of the hydrogen atoms as two point charges kept at a distance of 5.3 × 10–11 m from each other in vacuum. Also, we know that
Mass of an electron = me = 9.1 × 10-31 Kg
Mass of a proton = mp = 1.67 × 10-27 Kg
Charge of an electron = qe = - 1.6 × 10-19 C
Charge of a proton = qp = +1.6 × 10-19 C
By Universal Law of Gravitation,
By Coulomb’s Law,
It is noteworthy that the ratio is independent of the separation r between the charged particles. Also, there is a massive difference between the magnitudes of gravitational force and electrostatic force. The electrostatic force between the electron and the proton of the hydrogen atom is around 1039 times stronger than the gravitational force between them. Thus, from now onwards we will consciously neglect the gravitational force of attraction between charged particles.
Q.4. Charges of magnitude 100 µC each are located in vacuum at the corners A, B and C of an equilateral triangle measuring 4m 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?
Ans. 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
Thus the two forces are equal in magnitude. The angle between them is 120º, by geometry. The resultant force F is given by the vector law of addition
Thus, by vector law of addition, Magnitude of
Note: Observe that this force is parallel to AB.