1. COULOMB’S LAW IN
ELECTROSATICS
It states that electrostatic force of attraction or
repulsion between two point charges is directly
proportional to product of magnitude of charges
and inversely proportional to the square of the
distance between the two charges and it acts along
the line joining two charges.
Point charge- If size of charged bodies are much
smaller than the distance separating them , the size
may be ignored and the charged bodies are treated
as point charges

3. If q1 and q2 and are two point charges separated by a
distance r in vacuum, then magnitude of force is given
by ,
F= k
𝑞1
𝑞2
𝑟2
where k is constant of proportionality and is given by
the equation, k = 1/4πε0 =9 x 109 N m2 C-2 .
Where ε0 is called the permittivity of free space.
The value of ε0 in SI unit is ε0= 8.854 x 10-12 C2 N-1 m-2 .
Fvacuum =
1
4πε0
𝑞1
𝑞2
𝑟2

4. Definition of 1 C
By Coulomb’s law F= k
𝑞1
𝑞2
𝑟2
If q1=q2 = 1C and r =1 m in vacuum then,
F=k=9 x 109 N.
1 C is magnitude that charge which get repelled by
an electric force of 9 x 109 N from an identical
charge placed 1m apart in vacuum.

5. Dielectric constant(K) or Relative
permittivity (εr)
It is defined as the ratio of permittivity of a
medium(ε) to permittivity of free space (ε0).
K= εr =
ε
ε0
It is a dimensionless or unitless physical
quantity.
K=1 for air or vacuum.
K= ∞ for good conductors like metals
K >1 for all other materials.

6. If q1 and q2 and are two point charges
separated by a distance r in a medium, then
magnitude of force is given by ,
Fmedium =
1
4πKε0
𝑞1 𝑞2
𝑟2
But, Fvacuum =
1
4πε0
𝑞1
𝑞2
𝑟2
i.e., K =
𝐹𝑣𝑎𝑐𝑢𝑢𝑚
𝐹𝑚𝑒𝑑𝑖𝑢𝑚
Dielectric constant or relative permittivity of a medium is
defined as the ratio of force between two charges in
vacuum to the force between the same two charges placed
at same distance apart in that medium.

7. Direction of force between charges:
 If q1q2>0,then they repel each other.
 If q1q2 <0,then they attract each other.

8. Coulomb’s law in Vector form:

9.  Then force on q1 due to q2 is denoted as 𝐹12,
𝐹12 = k
𝑞1
𝑞2
𝑟12
2 𝑟12 = k
𝑞1
𝑞2
𝑟12
3 𝑟12
 Similarly force on q2 due to q1 is denoted as 𝐹21,
𝐹21 = k
𝑞1
𝑞2
𝑟21
2 𝑟21= k
𝑞1
𝑞2
𝑟21
3 𝑟21
But 𝑟12 = 𝑟21
and 𝑟12 = - 𝑟21
Therefore 𝐹12 = k
𝑞1
𝑞2
𝑟21
2 (- 𝑟21 ) = - 𝐹21
i.e. Coulomb’s law in agreement with Newton’s third
law of motion.

10. Force between multiple charges:
 Principle of superposition: It states that when a charge is
acted upon by many forces due to neighbouring charges
then the net force acting on it is the vector sum of forces
due to individual charges.
Consider a system of three charges as shown in fig.

11.  Then force on q1 due to q2 is 𝐹12,
𝐹12 = k
𝑞1
𝑞2
𝑟12
2 𝑟12
 And force on q1 due to q3 is 𝐹13,
𝐹13 = k
𝑞1
𝑞3
𝑟13
2 𝑟13
Therefore the net force 𝐹1 on q1 two charges q2 and q3
is,
𝐹1= 𝐹12 + 𝐹13
= k
𝑞1
𝑞2
𝑟12
2 𝑟12 + k
𝑞1
𝑞3
𝑟13
2 𝑟13

12.  If there is a system of n charges then the net
force on q1 is:
𝐹1= 𝐹12 + 𝐹13 + 𝐹14 +-------+ 𝐹1𝑛
= k
𝑞1
𝑞2
𝑟12
2 𝑟12 + k
𝑞1
𝑞3
𝑟13
2 𝑟13+-------+ k
𝑞1
𝑞𝑛
𝑟1𝑛
2 𝑟1𝑛
= kq1 𝑖=2
𝑛 qi
𝑟1𝑖
2 𝑟1𝑖

13.  Read NCERT text page no: 10 – 17
 Example 1.6 and 1.7
 Exercise 1.1,1.2 ,1.6,1.12 and 1.13
Assignments
1. Four point charges q ,Q , q and Q are placed at the
corners of a square of side a as shown in fig. Find
resultant electric force on charge Q.(CBSE 2018
Foreign qn)

14. 2. Three point charges q,-4q and 2q are placed at the
vertices of an equilateral triangle of side l as shown
in figure. Obtain an expression for the magnitude of
electric force acting on charge q.(CBSE 2018 Foreign
qn)