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**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 q_{1} and q_{2}, 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 q_{1} charge due to q_{2} beand force on q_{2} charge due to q_{1} charge be The corresponding vector from q_{1} to q_{2} is

**Note that by triangular law of vector addition: **

Similarly, vector leading from q_{2} to q_{1} 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 q_{1} and q_{2} when force is acting on q_{2} due to q_{1} is expressed as:

Therefore,

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 q_{1} and q_{2} will be:

**Case (a):** If q_{1} and q_{2} are same type of charges i.e. q_{1} > 0, q_{2} > 0 or q_{1} < 0, q_{2} < 0 then the electrostatic force between q_{1} and q_{2} will be repulsive.

**Case (b):** If q_{1} and q_{2} are opposite types of charges i.e. q_{1} > 0, q_{2} < 0 or q_{1 }< 0, q_{2} > 0 then the electrostatic force between q_{1} and q_{2} 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 q_{1}, q_{2}, q_{3}, q_{4}, then the resultant force exerted by charge 2, 3 and 4 on charge 1 is

Where is the net electrostatic force acting on charge q_{1}is the electrostatic force on q_{1} due to charge q_{2}, is the electrostatic force on q_{1} due to charge q_{3} and is the electrostatic force on q_{1} due to charge q_{4}.

**SOLVED EXAMPLES****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 q_{1} = 1C, q_{2} = – 3C and r = 3m. Then using Coulomb’s Law and substituting above values we get

Substituting the given values,

F = 3 × 10^{9} 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

q_{1} = charge of first body = q_{2} = charge of second body = q

r = 1 cm

Use the Coulomb’s Law equation

Substituting the values,

Thus,

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

Mass of an electron = m

Mass of a proton = m

Charge of an electron = q

Charge of a proton = q

Thus,

By Universal Law of Gravitation,

By Coulomb’s Law,

Thus,

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 10

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

Solving,**Note:** Observe that this force is parallel to AB.

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