# Electrostatics - 2

**Illustration: **

Q-1: A field of 10^{3} N/C acts at a point. What will be the force on a charge of 3´10^{-6}C and -2´10^{-6}C, 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.