VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
45 Questions around this concept.
The electric field inside a spherical shell of uniform surface charge density is
The electric field near a conducting surface having a uniform surface charge density
What is Volume charge distribution for Non conducting charged sphere
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Two infinite planes each with uniform surface charge density
There are two metallic spheres of the same radii but one is solid and the other is hollow, then
A charge of Q coulomb is placed on a solid piece of metal of irregular shape. The charge will distribute itself _____.
The figure shows a charged conductor resting on an insulating stand. If at the point P, the charge density is
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
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A circular metalling ring of radius R is given a charge +Q which is uniformly distribution on the circumference of the ring:
If a charged particle of charge Q and mass m enter perpendicularly in uniform electric field
Discrete charge distribution: A system consisting of many individual charges.
Continuous charge distribution: An amount of charge distributed uniformly or non uniformly on a body. It is of three types -
1. Linear charge distribution:
Example: wire, circulating ring
2. Surface charge distribution:
Example: plane sheet
3. Volume Charge distribution
Example - charge on a dielectric sphere etc.
Now we will discuss one example and derivation of the electric field due to uniformly charged rod.
So, let us consider a rod of length which has uniformly positive charge per unit length lying on x-axis, dx is the length of one small section. This rod is having a total charge Q and dq is the charge on dx segment. The charge per unit length of the rod is
. We have to calculate the electric field at a point P which is located along the axis of the rod at a distance of 'a' from the nearest end of Rod as shown in the figure -
and can be integrated because all electric field lies in the same direction.
Now, here
The total lield at
If
Now if we slide the rod toward the origin and the
- Electric field strength due to a charged circular arc at its centre
Let's try to find out the electric field at the center of an arc of linear charge density
If Q is the total charge contained in the arc then,
By symmetry, we know that the electric field due to the arc will be radially outward at the center.
Now consider a small element of the arc of charge
Now resolving the electric field due to small element of the
we see that both the vertical components get canceled and all that remains will be the horizontal component of the electric field due to the corresponding small elements.
So we have
so, by integrating all those electric fields due to these small elements we get the electric field due to the whole arc.
We know that electric due to charge dQ is given by,
But,
Substituting (2) in (1)
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