Illustration:
Q-1: A field of 10³ N/C acts at a point. What will be the force on a charge of 3´10^-6C and -2´10^-6C, kept there?
solution:

For +3MC

For -2MC       

Discussion:
The direction of force changes for different charges. For a positive field (due to +ve charge)

1)      The force is in direction of field on the +ve charge i.e. repulsive.

2)      The force is opposite to field direction on the –ve charge i.e. attractive.
Similar inference can be drawn for negative charges.
Electric field due to:

a)  Charged ring of charge Q and radius R.

1) The Center=0.
Derivation:

Linear charge density (l) on ring =

Consider the field at center due to any element =

But the field due to point diametrically opposite = in opposite direction.
Net field at center = 0 (By symmetry)

2)      On the axis =

On axis of ring at distance x.
Derivation:
         
                                  Fig (5)

As obvious from the diagram the field component along the line gets added due to opposite element.

*) By substituting x=0 in 2^nd result, we can get the first result.

*) Student should verify that the graph is of the following manner.

Fig (6)

b)      Due to a straight charged rod of length 2L with charge per unit length ‘’ at a distance ‘a’ on its perpendicular bisector.

E =


Derivation:

The rod is divided into infinitely small elements and the field due to symmetrically; opposite part add up as shown in figure (7).
Net field at P.
E =
dE =

               
                                      Fig (7)



Useful Tips:
1) If x>>a, E =  like a point charge.
2)  If L 
i.e. for infinite length

Question:

1)      How did the electric field cancel in one direction and add in another?

Ans:
                  
Observe the direction of electric field of two points which are symmetrically opposite.
Along the axis, net field =

Along the perpendicular,
Net field =
Similarly all fields along the perpendicular to axis cancel out.