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UNIT-1
Electrostatics
Structure of the Unit
1.0 Objectives
1.1 Introduction
1.2 Electric Field
1.3 Gauss’s Law
1.4 Illustrative Examples
1.5 Self Learning Exercise-I
1.6 Scalar Potential
1.7 Electrostatic Boundary Conditions
1.8 Discontinuity in Potential due to Dipole Layer
1.9 Illustrative Examples
1.10 Self Learning Exercise-II
1.11 Summary
1.12 Glossary
1.13 Answers to Self Learning Exercises
1.14 Exercise
1.15 Answers to Exercise
References and Suggested Readings
1.0Objectives
1.0 Objectives
This unit constitutes the basic concepts of static (time invariant )electric
field and potential. One can learn the Gauss’s law and its applications. We learn
the usefulness of spherical and cylindrical coordinates for solving the certain kinds
of problems in electrostatics. The better physical insight of behaviour of electric
field and potential across the interface can be got by studying the boundary
conditions at the interface.

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,1.1Introduction
1.1 Introduction
This unit introduces Coulomb’s law ,Gauss’s Law and boundary conditions
on electric fields and potentials. Gauss’s law is developed and shown in both
integral and differential form. The concept of circulation of the electric field is
related to its conservative nature is discussed. The concept of boundary conditions
for electric field and potential is introduced in this unit.
1.2
1.2Electric
ElectricField
Field
According to Coulomb’s law , electrostatic force F between two point charges q
and Q which are placed in free space at a distance r is expressed mathematically
as
Qq
F k
r2
This force acts along the line joining the charges.
2
1 9 N m
k  9  10
4 0 C2

12 C2
 0  8.854  10
N m2
The constant  0 is called permittivity of free space.
Force on q2 due to q1 can be written in vector form as
 1 q1q2 ˆ 1 q1q2 
F r  r (1)
4 0 r 2 4 0 r 3


.
q2

r12

r2 . q1


r1

O rig in

Figure1.1

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,  1 q1q2 ˆ 1 q1q2 
 F12  r  r12
4 0 r122 4 0 r132
12


 1 q1q2  
 F12   r2  r1 
4 0 r2  r1 3

  
where r12   r2  r1  =Position vector of q2 −Position vector of q1
  
r12  r2  r1
From eq.(1) taking q1  Q (Source charge) and q2  q (test charge),we have
 1 qQ ˆ
F r
4 0 r 2
  1 Q 
F  q rˆ 
 4 0 r 2

 
F  qE (Force on the test charge q)
 1 Q ˆ
E r (2)
4 0 r 2

 F
and E
q

E is called electric field (or electric field intensity) due to point charge Q.

SI unit of E is N .
C

Thus “Electric field E at a point is the force experienced per unit charge at rest
 
state at that point of space.” E  E (r ) has a value at each point in the space, so it is
called vector point field.
For definition of electric field , we can write

 F
E  lim
q0 q


The test charge q should be infinitesimally small because large value of the test
charge will disturb the original charge distribution of primary charges that

produces E .

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, If there are more than two charges, then we use principle of superposition for
determination of the force on a particular charge. If there are N point charges
Q1 , Q2 , Q3 ,...QN (source charges) placed respectively at distances
r 1 , r 2 , r 3 ,... r N from charge q then by principle of superposition total force on the
charge q(test charge)is given by
    
F  F1  F2  F3  ... FN
 qQ qQ qQ
F  k 21 rˆ1  k 22 rˆ 2  .... k 2N rˆ N
r1 r2 rN
   
Here position vectors of Q1 , Q2 , Q3 ,...QN are r1 , r2 , r3 ,... rN respectively and test

charge q is located at position r , then
     
r 1  r  r1 , r 2  r  r2 ,....
  Q Q Q 
 F  q  k 12 rˆ1  k 22 rˆ 2  .... k 2N rˆ N 
 r1 r2 r N 

 
 F  qE
 Q Q Q
where E  k 12 rˆ1  k 22 rˆ 2  .... k 2N rˆ N
r1 r2 rN
 Q
E  k  2i rˆ i
ri
    
or E  E1  E2  E3  ... E N
Above expression represents the principle of superposition for electric field.
For continuous charge distribution
 1 dq
E rˆ
4 0  r 2

Now we consider the Coulomb’s law for the general case of volume charge.
Volume charge density is   dq (in C/m3) ,where differential charge dq is
d
present in a differential volume d

The electric field at a point r   x, y, z 
in terms of integral over the volume
charge distribution   x, y, z is written as
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