Electric Field IntensityIt is the force experienced by a unit positive charge placed at a point in an electric field.

Due to a point charge

Due to linear distribution of charge

λ = linear charge density of rod

(i) At a point on its axis.

(ii) At a point on the line perpendicular to one end

Due to ring of uniform charge distribution

At a point on its axis

Due to uniformly charged disc.

At a point on its axis.Thin spherical shell

1. Non conducting solid sphere having uniform volume distribution of charge

(i) Outside point(ii) Inside point

q = total charge

ρ = volume density of charge

Cylindrical Conductor of Infinite length

Outside pointInside point λ = linear density of charge

E = 0 (as charges reside only on the surface)

Non-conducting cylinder having uniform volume density of charge

Infinite plane sheet of charge

λ = surface charge density

Two oppositely charged sheets (Infinite)

(i) Electric field at points outside the charged sheets

EP = ER = 0

(ii) Electric field at point in between the charged sheetsExample 3. Find electric field intensity due to long uniformly charged wire.

(charge per unit length is λ)

Example 4. What is the electric field at any point on the axis of a charged rod of length `L' and linear charge density`λ'? The point is separated from the nearer end by a.

Solution : Consider an element, dx at a distance x from the point, P, where we seek to find the electric field. The elemental charge, dq =λdx

Example 5. A ring shaped conductor with radius R carries a total charge q uniformly distributed around it as shown in the figure. Find the electric field at a point P that lies on the axis of the ring at a distance x from its centre.

Solution: Consider a differential element of the ring of length ds. Charge on this element is

This element sets up a differential electric field d at point P.

The resultant field E¯ at P is found by integrating the effects of all the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of d E¯ parallel to this axis contributes to the final result.To find the total x-component E

_{x}of the field at P, we integrate this expression over all segment of the ring.

The integral is simply the circumference of the ring = 2πR

As q is positive charge, field is directed away from the centre of the ring, along its axis.