Question

The electric field intensity due to hollow charge at any point is:

Answer 1

Hint: We know that the force is the result of repulsive force from a similar charge present on the rest of the surface of the conductor. Let’s consider a small element $ds$ on the surface of a charge conductor. If $\sigma $ is the surface density and the charge carried by the element $ds.$

Complete answer:
Electric field intensity can be determined by finding the strength of the electric field at that point. For calculating electric force and electric pressure, we will use the relation between electric field vectors and the surface density of a conductor. Electric field intensity is expressed as the strength of an electric field at any point. This value is equal to the electric force per unit charge experienced by a test charge placed at that point. Every element of a charged conductor experiences some normal mechanical force.

When the body is a spherically symmetric shell, like a hollow sphere, there will be no resultant gravitational force acted by the shell on any of the bodies inside this without considering the location of the object within the shell. The intensity of the electric field inside a uniformly charged hollow sphere will be zero and also, we can say that outside the sphere, we can suppose that the sphere will be a point charge present at center and can be calculated by the electric field. Thus, the potential will also be a constant inside this hollow sphere as the electric field becomes zero.

Therefore, the electric field intensity due to hollow charge at any point is zero at any point inside the sphere.

Note: Remember that while calculating the electric field at any point, the direction of the vector should be considered very carefully. Also remember that the electric field inside a conductor is zero because free charges in a conductor reside only on the surface.