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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
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
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
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 sheets
Example 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 Ex 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.