Question

A spherical cavity of radius R/2 is made inside a fixed uniformly charged non-conducting sphere of radius R and volume charge density rho as shown.A particle of mass m and charge - q is released from rest from centre of cavity. Time taken by it to hit the surface of cavity is​

Answer 1

Answer:

The problem can be considered as a superposition of electric field E due to the larger sphere without the cavity and a smaller sphere with opposite charge density.

Inside a spherical charge distribution for an internal point 

E(r)=3ϵrρr  which is a result of the application of Gauss's law.

Let the vector to the point P inside the cavity be r1  taking origin at the center of the large sphere and  r2 be the vector from center O' of the smaller sphere.

∴E=3ϵ0ρr1+(3ϵ0−ρr2)=3ϵ0ρ(r1−r2)=3ϵ0ρa where a is the vector joining the centres O and O' of the large and small spheres which is a constant vector.

∴ electric field is uniform both in magnitude and direction.

Answer 2

Answer:

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