4. SUPERPOSITION APPLIED TO CONDUCTORS a) Consider a solid conducting sphere wit

Question

4. SUPERPOSITION APPLIED TO CONDUCTORS a) Consider a solid conducting sphere with radius a and a total charge +q0. Calculate and plot the electric field as a function of radial distance r. b) Consider separately a conducting spherical shell with inner radius b and outer radius c and total charge -q0. Calculate and plot the electric field as a function of radial distance s. c) Now consider the two conductors from parts (a) and (b) arranged concentrically (where b > a). Use the principle of superposition to calculate the field in all space, including the regions inside the conductors. This will lead to an inconsistency. Have you applied superposition correctly? Is Gauss' law invalid here? How does nature resolve this inconsistency?

Explanation / Answer

Given there is a solid conducting sphere of radius a and charge = q0

Since the sphere is conducting, all the charge exists on the surface.

So Electic field inside the sphere at any point is always zero.

The field on the surface and outside can be discovered by using gauss law.

integral of ( E.ds) = q in/ epsilon 0.

So E is constant on any point of surface, but the direction varies. This surface is a guassian surface which is spherical in this problem and coincides with the solid sphere surface in this problem.

So E = q/(4* pi* * r^2 *epsilon) where r varies from "a" to infinity.

(b)

Now in second case, the conducting shell again where the charge stays on the outside.

So Electric field inside the hollow space is again zero at every point.

This is because of symmetry but not due to guass law.

Guass law tells always integral (E.ds) = q in /epsilon 0 But not E =0

So if q in is zero implies the whole integral is zero. Not the electric field.

THere are cases where E exists but the surface integral on guassian surface becomes zero.

Another notable thing is Electric field inside a conductor is zero. This is not by guass law. But material property itself.

All conductors conduct electricity only when there is a potential difference exists. So if there is charge inside a conductor, there must be some potential difference so cureent flows until potential difference is zero. Since all the charge distribution tries to minimize its potential energy of the system by nature, everything repels and comes to surface. So no charge ever stays inside a conductor unless potential difference exists.

So now qustion is what causes charge to repel and rearrange configuration. Its the field. So where there is field there is force. So until, the force inside a conductor becomes zero, the charges reconfigure.

So E itself is zero so q in has to be zero. this is logic in conductor. Both E and q in goes to zero.

In case B in hollow region, there is no conducting material and vaccum. And no charge as well. So q in =0 . But why is V zero. Let us take a guassian circular surface and at every point on surface there exists some field may be with different magnitutde and different direction altogether.

But since the charge distribution outside is symmetrical, E at every point on circle should make same angle with radius and magnitude must be equal. ( this is due to the symmetrical charge distribution in this case. Not due to spherical guassian surface. Had the distribution been uneven, even taking spherical surface may yield different magnitude and direction on guassian surface)

Noe dot product is the component of Electric field along radius. Since all the magnitudes and directions are same and adding them vectorially it would be a zero. That is why E in hollow space is zero. So q in automatically becomes zero.

now in case (c)

Let us say , sphere is put incide the shell. Now by same arguments. E inside the conductor from 0 to a is zero. and E inside the conducting shell b to C is zero. Also outside C it is same as net q inside the guassian surface. Since everything is sphere net q is =q and -q which is zero.

What happens between A and B.

The answer is there exists field in hollow space where q in exists as +q on the surface of solid sphere a.

So E .ds = q in /epsilon 0. Note that guass law says E is the net field at a point due to all charge configuration not just single charge.

SO if we divide this into 2 parts E on solid and E on outside spherical. So From outside shell the contribution to hollow space is zero as we have discussed due to symmetry of field and integral becoming zero.

Now between a and b +q conributes showing E = (1/4*pi * epsilon * r^2)

Another question is if there exists a field why does not the charge move from solid sphere to vacuum to outside conducting sphere?

As we have already told, the force exerted by outside shell is zero in its hollow space. So what everchrge is inside does not experience any thing and stays there whereever it is put. Since this inside charge can create any electric field that is felt in the space. THis also creates induced charge on surfaces of outer shell which then reconfigures again making the guass law righr arranging the net electric field direction accordingly.

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