# Pls give a solution of it as i am unable to get the write ans this question

Arun
25758 Points
2 years ago
Dear student

Hope you understand

Thanks and regards
Arun
Parth
58 Points
2 years ago
In this case, you’ve to take spherical element at distance r and using gauss theorem, you’ll find that for r1
E due to charge distribution  = k*q*(1/(r1)^2 – 1/(r)^2) so when you add the electric field due to charge inside cavity the net E will remain constant for that region. For rr2 it’ll be acting like an E due to charge particle
hence option B
pls approve if it helps
Ayush_Deep
120 Points
2 years ago
Friend , the image is clear .
The correct option will be option B.
For any point outside the sphere , X>R2 . The whole charge can be assumed to be at centre and then electric field at an outer point will be kQnet /r². And the graph will be hyperbolic for distance greater than R2.
For X less than R2 but greater than R1,
The charge density is not uniform and density = dQ/dV
dV is 4πr² Dr.
By this you you can find equation of charge varying with radius Q directly proportional to r².
The electric field is kq/r² and will be constant for this region.
Now third case , for an inside point apply guass law  the charge is kept at centre due to which the graph again be hyperbolic.
Hope you understood.kindly approve