We all know that; electrostatics is the branch of physics in which electric charges which are at rest can be studied. When somebody is rubbed with other body then there will be transfer of some electrons from the atom of one body to another takes place. And the body which receives electrons will become negatively charged and the body who lose electrons will become positively charged. So electric charges are of two types one is negatively charged called as electrons and other positively charged. The particles with no charge are the neutral particles.
We all know that, like charges repels each other and unlike charges attract each other.
This behaviour of like and unlike charges with each other is explained by the scientist Coulomb in 1785 which is stated as follows.
Statement:
According to Coulomb’s law, the electrostatic force of attraction or repulsion between any two electric charges which may be like or unlike but they are at rest is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between that two electric charges. And this electrostatic force of attraction or repulsion between these two electric charges is acting along the line joining the charges.
Explanation:
If we have considered two electric charges having charge q1 and q2 and they are separated by distance r then electrostatic force of attraction or repulsion between these two-point charges which are at rest is given by,
F α q1*q2
F α 1/r2
Thus, F α q1*q2/r2
And F = K q1*q2/r2
Where K is the constant of proportionality and if charges are in vacuum then value of K is taken as K = 1/4πɛ
And ɛ is the permittivity of the free space.
So, we can write Coulomb’s law in vacuum also as given below,
F = 1/4πɛ q1*q2/r2
In vector form:
The following diagram shows the vector form of Coulomb’s law of electrostatics.
Here, q1 and q2 are the like charges separated by distance r.
Then the vector form of Coulomb’s force exerted by charge q1on the charge q2 is given by,
F21 = 1/4πɛk(q1*q2/r2) r12
Where r12 is the unit vector whose direction is from charge q1 to q2.
In similar manner, the vector form of Coulomb’s electrostatic force exerted by charge q2 on charge q1 is given as,
F12= 1/4πɛk(q1*q2/r2) r21
Where r21is the unit vector whose direction is from q2 to q1.
From vector algebra, we write as r12= –r21
Thus, we also write F21= –F12
Example:
If the two like charges having magnitude of charge 5 C are separated by the distance 3m then what will be the electrostatic force of repulsion between them.
Solution:
Given that, q1= 5C
And q2 = 5 C
And r = 3m
We know that, 1/4πɛ = 9*109Nm2/C2
Thus, electrostatic force of repulsion is given by
F = 1/4πɛ q1*q2/r2
F = 9*109*5*5/ 9
F= 25*109 N
Thus, the electrostatic force of repulsion between the given two like charges is found to be 25*109 N.
Note:
- The Coulomb’s law of electrostatics is only applicable to the electric charges which are at rest.
- It is applicable to only point charges and not for the other extended bodies also.