Electric Field Intensity

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

Due to a point charge

 1761_electric field.JPG

Due to linear distribution of charge

        λ = linear charge density of rod

 

2200_electric field.JPG

(i)    At a point on its axis.

 855_electric field.JPG

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

1644_electric field.JPG

Due to ring of uniform charge distribution

At a point on its axis

 

980_electric field.JPG

Due to uniformly charged disc.
At a point on its axis.

159_electric field.JPG

Thin spherical shell

 1940_electric field.JPG

1.    Non conducting solid sphere having uniform volume distribution of charge
(i)    Outside point

2082_electric field.JPG

(ii)    Inside point    
           q = total charge
           ρ = volume density of charge

 2487_electric field.JPG

Cylindrical Conductor of Infinite length

Outside point

1594_electric field.JPG

Inside point                    λ = linear density of charge
    E = 0        (as charges reside only on the surface)
Non-conducting cylinder having uniform volume density of charge

 1566_electric field.JPG

Infinite plane sheet of charge

 1424_electric field.JPG

 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 sheets

308_electric field.JPG

Example 3.    Find electric field intensity due to long uniformly charged wire. 
    (charge per unit length is λ)

 143_electric field.JPG

 

322_electric field.JPG

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.

 

2178_electric field.JPG

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

 

1225_electric field.JPG

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.

 

2070_electric field.JPG

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

 1771_electric field.JPG

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.

1466_electric field.JPG

To find the total x-component Ex of the field at P,  we integrate this expression over all segment of the ring.

 15_electric field.JPG

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

 610_electric field.JPG

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

 
Related Resources
Gauss Theorem and Examples

GAUSS THEOREM The flux of an electric field...

Capacitor and Combinations

CAPACITORS Capacitance If Q is the charge given to...

Dipole

DIPOLE The term electric dipole stands for two...

Coulombs Law

Coulomb’s Law:- In 1785 Charles Coulomb...

Capacitors and Energy Interpretation

Energy Stored in a Capacitor A capacitor is...

Electric Potential and Energy

Electric Potential The electric potential is a...