Charge in a conducting spherical shell ?
A conducting spherical shell of inner radius a and outer radius b carries a net charge Q. A point charge q is placed at the center of this shell. Determine the surface charge density on (a) the inner surface of the shell and (b) the outer surface of the shell.
A conducting spherical shell of inner radius a and outer radius b carries a net charge Q. A point charge q is placed at the center of this shell. Determine the surface charge density on (a) the inner surface of the shell and (b) the outer surface of the shell.
Inner Surface : σa=qa4πa2=−q4πa2
Outer Surface : σb=qb4πb2=Q+q4πb2
Explanation :
Since the electric field must necessarily vanish inside the volume of the conducting sphere, the charges must drift in such a way as to cancel the electric field due to the charge q at the centre.Inner Surface: Consider an imaginary sphere enclosing the inner surface of radius a, lying just outside this surface and inside the volume of the conducting sphere. By Gauss's Law the electric flux through this surface is related to the total charge enclosed by this surface,
By Gauss Law, ΦE=∮→E⋅→ds=q+qaε0
where q is the charge at the centre and aa is the total induced charge on the inner surface.
Since the Electric field vanishes everywhere inside the volume of a good conductor, its value is zero everywhere on the Gaussian surface we have considered. So the surface integral is zero.
ΦE=∮→E⋅→ds=q+qaε0=0; →qa=−q ...........(1)
This is the total charge induced on the inner surface. Because the electric field from the centra;l charge is spherically symmetric, this induced charge must be distributed uniformly distributed too.
So the charge density on the inner sphere is: σa=qa4πa2=−q4πa2
Outer Surface: The net charge on the outer surface has two components - free charge qfreeb=Q and induced charge qindb
qb=qindb+qfreeb=qindb+Q
Because the induced charges are a result of polarization due to the electric field of the central charge, the net induced charge on the inner and outer surfaces of the good conductor must be zero :
qa+qindb=0;qindb=−qa
Writing qa in terms of q using (1) , qindb=−qa=q
Thus the total charge on the outer surface is: qb=Q+b
So the charge density on the outer sphere is: σb=qb4πb2=Q+q4πb2
Outer Surface : σb=qb4πb2=Q+q4πb2
Explanation :
Since the electric field must necessarily vanish inside the volume of the conducting sphere, the charges must drift in such a way as to cancel the electric field due to the charge q at the centre.
Inner Surface: Consider an imaginary sphere enclosing the inner surface of radius a, lying just outside this surface and inside the volume of the conducting sphere. By Gauss's Law the electric flux through this surface is related to the total charge enclosed by this surface,
By Gauss Law, ΦE=∮→E⋅→ds=q+qaε0
where q is the charge at the centre and aa is the total induced charge on the inner surface.
Since the Electric field vanishes everywhere inside the volume of a good conductor, its value is zero everywhere on the Gaussian surface we have considered. So the surface integral is zero.
ΦE=∮→E⋅→ds=q+qaε0=0; →qa=−q ...........(1)
This is the total charge induced on the inner surface. Because the electric field from the centra;l charge is spherically symmetric, this induced charge must be distributed uniformly distributed too.
So the charge density on the inner sphere is: σa=qa4πa2=−q4πa2
Outer Surface: The net charge on the outer surface has two components - free charge qfreeb=Q and induced charge qindb
qb=qindb+qfreeb=qindb+Q
Because the induced charges are a result of polarization due to the electric field of the central charge, the net induced charge on the inner and outer surfaces of the good conductor must be zero :
qa+qindb=0;qindb=−qa
Writing qa in terms of q using (1) , qindb=−qa=q
Thus the total charge on the outer surface is: qb=Q+b
So the charge density on the outer sphere is: σb=qb4πb2=Q+q4πb2
