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I need help understanding how induction works with a spherical conductor. Say there is a spherical conductor having a spherical cavity and a positive charge in the center of the cavity. Negative charges are induced on the inner surface of the sphere and equal positive charges are induced on the outer surface. My worry is if we consider the two surfaces of the sphere as one inner and one outer ,each surface will have two surfaces one inner and one outer. Why are four forms of inductions are not taken? I.e., negative charge on the inner surface of the inner sphere, positive charges on the outer surface of the inner sphere, negative charges on the inner surface of the outer sphere and positive charges on the outer surface of the outer sphere.

If we take two surfaces only, why does the potential of a point lying outside the sphere depend only the induced positive charges of the outer sphere? The book says the electric field at the point is zero due to positive charge in the cavity and the induced negative charge on the inner surfaces. How can this be possible?

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  • $\begingroup$ It does not happen because if it did, it would violate the Gauss' Law. In other words, if charges were induced in the inner surfaces of the two surfaces, then there would be a net field inside the conductor which would drive a current. $\endgroup$ – Yashas Mar 17 '17 at 13:14
  • $\begingroup$ a reverse question still persists,there are some positive charges lying on the outer surface and some negative charges lying in the inner surface which will result in a potential difference and a current may flow,pl clarify. – sachinrath123 14 mins ago $\endgroup$ – sachinrath123 Mar 17 '17 at 14:14
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How do you make out that there are 4 surfaces? Each surface is the boundary between air and metal. The conductor has only 2 boundaries. Excess charge exists only on the air side of each of these boundaries.

There are several ways of looking at the electric field outside the conductor.

  1. No electric field escapes from inside the conductor. Every line of the electric field starts on the +ve charge at the centre and ends on one of the induced -ve charges on the inner surface. This is the same as for ideal spherical or parallel plate capacitors : the charge on one plate is exactly matched by opposite charge on the other plate, so there is no electric field outside of the capacitor.

The only unmatched charge is the +ve charge induced on the outer surface of the conducting sphere. Because this charge distribution is spherically symmetric, the field outside the sphere is the same as though this +ve charge were concentrated at the centre - just like the +ve charge which is actually there. See the Shell Theorem.

  1. The conductor is invisible; it produces no net field. The charges on the inner and outer surfaces of the conducting sphere are matched. All lines of electric field start on the +ve charges on the outer surface and end on an equal number of -ve charges on the inner surface.

The only unmatched charge is the +ve charge at the centre. The electric field at P is caused solely by this charge. It is as though the conductor is invisible.

  1. The field at any point is the superposition of 3 charge distributions. The field due to the point charge is radially outward with strength $E_1=+kQ/r^2$. There is no electric field inside the charge distributions on the inner and outer surfaces of the conductor because they are spherically symmetric (Shell Theorem). Outside of the inner and outer surfaces respectively the electric fields are radial with strength $E_2=-kQ/r^2$ (inwards) and $E_3=+kQ/r^2$ (outwards). The total electric field outside of all 3 charge distributions is $E_1+E_2+E_3=+kQ/r^2$, the same as for the point charge alone.
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  • $\begingroup$ what could be the potential and electric field at any point inside the cavity,will they remain constant or vary ? $\endgroup$ – sachinrath123 Mar 18 '17 at 12:33
  • $\begingroup$ Inside the cavity the electric field is the same as for the isolated point charge, because the electric field inside a spherical charge distribution (outer and inner surfaces of the shell) is zero. The electric field outside the conducting shell is as usual, whereas inside the shell is zero. A +ve test charge approaching from infinity encounters the same $E$ field as for the isolated point charge, except inside the conducting shell, where it is zero. It requires less work to reach the cavity, so the electric potential inside the cavity is lower than for the isolated point charge. $\endgroup$ – sammy gerbil Mar 18 '17 at 16:28
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If you consider a gaussian surface inside the conducting material of the sphere, and the net charge inside the conductor is not $0$ then this would mean that there is electric field inside the conducting material, which doesn't agree with the experimental facts.

There is no electric field inside conducting material, because if there were, then there would be movement of charges, and this doesn't happen in the real world.

And for that to happen, there must be equal -ve charge in the inner surface of the cavity, so that whenever we consider a gaussian surface inside the conducting material the net charge becomes $0$.

The charges tend to occupy spaces in a conducting material where the repulsion between charges is minimum and hence, the stability is maximum.

The electric field lines due to the +ve charge inside the cavity never reach out of the sphere. But the field line due induced charge on the outer surface of the sphere are exactly like that of the +ve charge at the centre. So one can assume that the sphere acts like a charged conductor with a +ve charge at its centre. This can be said only for the fields outside the sphere. So if you calculate the electric field at the points outside the sphere, it can easily be calculate by assuming a charge at the centre with magnitude equal to the induced charge.

In the conducting material E.Field is $0$, so there will be no work done by external agent in moving a test charge here. So the potential here remains the same and equal to potential of the test charge at the outermost part of the sphere.

And now when you move inside the cavity, the electric field lines due to +ve induced charge do not reach inside. Its the -ve induced charge and the +ve charge whose field lines are existing inside the cavity. The contribution of the +ve induced charge ends at the outermost surface of the sphere. The potential, here, will be due to -ve induced charge and the +ve charged particle.

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