A hollow spherical conducting shell (in electrostatic equilibrium) has an inner

Question

A hollow spherical conducting shell (in electrostatic equilibrium) has an inner radius R1 and an outer radius R2 with a total charge -2Q distributed uniformly on its surfaces. the inside of the hollow spherical conducting shell is filled with nonconducting gel with a total charge of -3Q distributed as p = p_0*r (where p_0 is a constant) throughout the volume. Find the surface charge density induced on the surface of Radius "R1" Find the surface charge density induced on the surface of Radius "R" Express the constant p_0 in terms of Q and any other variables given. State the units of p_0

Explanation / Answer

as the shell is conducting, no charge can reside inside.

and electric field inside a conducting surface is 0.

as per gauss law, as electric field is directlly proportional to charge enclosed, zero electric field means

total charge enclosed by the hollow sphere is zero.

hence

charge on inner surface of the hollow sphere+ charge on the non conducting gel=0

==>charge on inner surface of hollow sphere=-(-3*Q)=3*Q

total charge on the hollow sphere =-2*Q

==>charge on outer surface+charge on inner surface=-2*Q

charge on outer surface=-5*Q

part a:

charge density on inner surface of radius R1=3*Q/(4*pi*R1^2)

part b:

charge density on outer surface of radius R2=-5*Q/(4*pi*R2^2)

part c:

consider a sphere of radius r such that r<R1.

if we consider charge enclosed between r and r+dr,

charge enclosed=4*pi*r^2*dr*(pho_0*r)

=4*pi*pho_0*r^3*dr

total charge=-3*Q

=>integration of 4*pi*pho_0*r^3*dr from 0 to R1=-3*Q

==>pi*pho_0*(R1^4-0^4)=-3*Q

==>pho_0=-3*Q/(pi*R1^4)

unit of pho_0 is C/m^4.

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