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

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(i)    At a point on its axis.

 855_electric field.JPG

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

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Due to ring of uniform charge distribution

At a point on its axis

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

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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

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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.


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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


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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.


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