Redistribution of the charges in three concentric spherical shells
When we bring a charged object near an uncharged object, an equal and opposite charge is induced on it.
Similarly when two charged bodies bring near to each other, some charge will also induce on them.
This induced charges causes the redistribution of the charges already present on the bodies.
A similar redistribution of the charges will occur on the surfaces of the charged concentric spherical shells.
Due to charge on one sphere, an equal and opposite charge will be induced on the other and vice-versa.
But before going into detail, let's see the concentric spherical shells.
The three concentric spherical shells.
This is the construction of three concentric spherical shells.
is the radius of inner sphere,
is the radius of middle sphere and
is the radius of outer most sphere.
The charge on the surface of the inner sphere is
and the charge on the surface of the outer sphere is
The middle shell is earthed.
Although, we have to find the redistribution of the charges on the different surfaces of the concentric spherical shell, we will have to keep in mind the properties of conductors.
Let's see some important properties of conductors.
The total charge on the isolated spherical shell remains constant.
The electric field inside the body of conductor is zero.
And the electric potential of the earthed surface is zero.
To find the redistribution of the charges we can use the Gauss Law.
It gives the relation between the electric field and and the distribution of the charges through the concept of Gaussian surface.
Let's find the redistribution of the charges.
Let's assume the Gaussian surface somewhere in the thickness of middle sphere.
Then the charge enclosed by the Gaussian surface will be the charge on the inner sphere and inner surface of the middle sphere.
Apply the Gaussian theorem to the assumed Gaussian surface in the second sphere.
As the charge enclosed is
, The expression will be
As we know that the electric field inside the charged conductor is
., the net charge should also be zero.
On proceeding the simplification, we get the charge on the inner surface of the middle shell equal to the
Similarly, take the outer sphere, and assume a Gaussian surface somewhere in the thickness of the outer shell.
Let's assume, charge induced at the inner surface of the outer shell is
Thus, the charge enclosed by the Gaussian surface will be as shown above.
Let's apply the Gauss theorem for the Gaussian surface considered on the outer shell.
On putting the value of the charge enclosed in the expression, we get
As the electric field inside the charged conductor will be
The net charge inside the conductor should also be zero.
On further simplification we will get,
The potential on the second or middle shell due to the charge on the first or inner shell will be given by,
The potential on the second or middle shell due to the charge on the middle shell will be given by,
And the potential on the second or middle shell due to the charge on the third or outer shell will be given by
Since, the second is earthed, its potential should be zero. Thus,
Thus, the potential of the second shell due to first, second and the third shell can be expressed as
On further simplification we get,
We obtained the charge on the outer surface of the inner shell.
So the distribution of the charge can be given as,
The electric field inside the charged conductor is
and the potential of the earthed conductor is
The charges before redistribution is
The charge on the surface of concentric shells after redistribution is