1) Point charges placed in space create electric fields that exert forces on other charges, either attracting or repelling them. 2) The electric field concept can explain how charges interact at a distance. 3) Coulomb's law describes the electrostatic force between two point charges quantitatively in terms of the charges and their distance.

2.  Consider a point charge kept at a point in
space. If another point charge is placed
at some distance from the first point
charge, it experiences either an attractive
force or repulsive force. This is called
‘action at a distance’.

3.  According to Faraday, every charge in the
universe creates an electric field in the
surrounding space, and if another charge is
brought into its field, it will interact with the
electric field at that point and will experience a
force.
 This field concept is required to explain action
at a distance

4.  Consider a source point charge q located at a
point in space.
 Another point charge qo (test charge) is placed
at some point P which is at a distance r from
the charge q.
 The electrostatic force experienced by the
charge qo due to q is given by Coulomb’s law.

5.  The charge q creates an electric field in the
surrounding space. The electric field at the point P
at a distance r from the point charge q is the force
experienced by a unit charge and is given by
 Here is the unit vector pointing from q to the point
of interest P. The electric field is a vector quantity
and its SI unit is Newton per Coulomb (NC-1).

8.  If the electric field at a point P is then the force
experienced by the test charge qo placed at
the point P is ,
 This is Coulomb’s law in terms of electric field.

11.  This equation implies that the electric field is
independent of the test charge qo and it
depends only on the source charge q.

12.  Since the electric field is a vector quantity, at
every point in space, this field has unique
direction and magnitude
 As distance increases, the electric field
decreases in magnitude.
 The strength or magnitude of the electric field
at point P is stronger than at the points Q and
R because the point P is closer to the source
charge.

15.  In the definition of electric field, it is
assumed that the test charge q0 is taken
sufficiently small, so that bringing this test
charge will not move the source charge.
 In other words, the test charge is made
sufficiently small such that it will not
modify the electric field of the source
charge

16.  This expression is valid only for point charges.
 For continuous and finite size charge
distributions, integration techniques must be
used

19.  Uniform electric field will have the same
direction and constant magnitude at all
points in space.
 Non-uniform electric field will have
different directions or different
magnitudes or both at different points in
space.
 The electric field created by a point
charge is basically a non uniform electric
field.

29.  To find the electric field at some point P due to
this collection of point charges, superposition
principle is used.
 The electric field at an arbitrary point due to a
collection of point charges is simply equal to
the vector sum of the electric fields created by
the individual point charges. This is called
superposition of electric fields

31.  Consider a collection of point charges located
at various points in space. The total electric
field at some point P due to all these n charges
is given by q1,q2,q3,------qn

39.  While dealing with the electric field due to a
charged sphere or a charged wire etc., it is
very difficult to look at individual charges in
these charged bodies.
 Therefore, it is assumed that charge is
distributed continuously on the charged bodies
and the discrete nature of charges is not
considered here.
 The electric field due to such continuous
charge distributions is found by invoking the
method of calculus.

40.  Consider the charged object of irregular shape

41.  The entire charged object is divided into a large
number of charge elements q1,q2,q3------
qn
 Each charge element q is taken as a point
charge
 The electric field at a point P due to a charged
object is approximately given by the sum of the
fields at P due to all such charge elements

43.  To incorporate the continuous distribution of
charge, we take the limit
 In this limit, the summation in the equation
becomes an integration and takes the following
form
 Here r is the distance of the point P from the
infinitesimal charge dq and r is the unit vector
from dq to point P

45.  If the charge Q is uniformly distributed along
the wire of length L, then linear charge density
(charge per unit length)isλ=Q/L.
 Its unit is coulomb per meter (Cm-1).
 The charge present in the infinitesimal length dl
is dq = λdl
 The electric field due to the line of total charge
Q is given by

47.  If the charge Q is uniformly distributed on a
surface of area A, then surface charge density
(charge per unit area) isσ=Q/A
 Its unit is coulomb per square meter (C m-2).
 The charge present in the infinitesimal area dA
is dq = σ dA
 The electric field due to a of total charge Q is
given by

49.  If the charge Q is uniformly distributed in a
volume V, then volume charge density (charge
per unit volume) is given byρ=Q/V
 Its unit is coulomb per cubic meter (C m-3).
 The charge present in the infinitesimal volume
element dV is dq = ρdV.
 The electric field due to a volume of total
charge Q is given by

56.  That the magnitude of the electric field is
directly proportional to the mass m and
inversely proportional to the charge q.
 It implies that, if the mass is increased by
keeping the charge constant, then a strong
electric field is required to stop the object from
sliding.
 If the charge is increased by keeping the mass
constant, then a weak electric field is sufficient
to stop the mass from sliding down the plane.
 The electric field also can be expressed in
terms of height and the length of the inclined
surface of the plane