Field due to internal Induced charge on a conductor to an external point?

A charge q is located at a distance r from the center of a conducting sphere with inner radius 2r. The charge induces charges on the inner surface of the sphere according to Gauss' law .

The electric field at point p is to be approximated.

Inside the material of the conducting sphere, the electric field due to induced charge will cancel out the electric field due to the charge inside the sphere. Accordingly the electric field lines will begin at induced charge and terminate at the inner charge.

Therefore the field due to internal induced charge on the point p must be zero , (note it may be nonzero due to external induced charge but the problem specifies internal)

The solution however says it to be $kq/17r^2$ and not zero Isn't the electrostatic system shielded from the conductor? • please reformulate your question to make it better understandable. Describe the shape of the sphere. where you expect an induced charge and what you mean by inner charge and where the charges are located. – Crimson May 9 '17 at 11:33
• @Crimson Is it okay now?please remove the downstate i am new to this site and thought my image would suffice – Ernest Earnshaw May 9 '17 at 11:54
• I edited your question (my edit needs to be approved first). Please check whether the edited question still conveys your question. Otherwise feel free to change it back. – Crimson May 9 '17 at 12:57

The electric field at the point P is solely due to the charges on the outer surface of the sphere [Suppose this was not true, for the sake of contradiction.The only way this can happen is if the magnitude of induced charge on the inner surface of the sphere is not equal to the q itself.If this happens, then by Gauss' law, we have a non-zero electric field in the meat of the conductor, which is impossible. This proves that the field at the point P is solely due to the charges on the outer surface of the sphere].

Thus, the the field due to the charge has to be cancelled by the charges induced on the inner surface. The induced charges have to be opposite to that of the charge q. The electric field due to the charge q at the point P from Coulomb's law is $kq/17r^2$. So, the field due to the induced charges on the inner surface has to be $kq/17r^2$ in magnitude but opposite in direction to that of the field direction due ti charge q alone.

There are three types of charges contributing to the field at point P

• The charge q itself
• The induced charge on the inside of the sphere
• The induced charge on the outside of the sphere.

First, the contribution due to the charge q itself will have a magnitude of $\frac{kq}{r^2}=\frac{kq}{(4R)^2+(1R)^2}=\frac{kq}{17R^2}$

Second, the field due to the induced charge on the inside of the sphere will cancel the field of the original charge and therefore have a magnitude which is equal but opposite to the field of the original charge.

Third, if the sphere is uncharged, i.e. electrically neutral, the accumulation of charge on the inside surface, will lead to an equally large, but opposite accumulation of charge on the outside surface. This charge will be evenly spread over the surface of the sphere. The magnitude of this field is $\frac{kq}{r^2}=\frac{kq}{16R^2}$

• the problem specifies internal induced charge so we cant account for the external induced charge – Ernest Earnshaw May 9 '17 at 13:02
• Kq/16r^2 is one of the option in this multiple choice question but is given incorrect by the solution key – Ernest Earnshaw May 9 '17 at 13:05
• to clarify the problem. I assume that we are talking about a perfectly conducting spherical shell with a finite thickness. The charge q is inside the shell and p is outside the shell. Is this correct? – Crimson May 9 '17 at 13:08
• @ErnestEarnshaw I updated my answer – Crimson May 9 '17 at 13:35
• I guess that solves my doubt – Ernest Earnshaw May 9 '17 at 14:34

Due to q charge on the sphere there will be no charge(or total charge is only at R distance from center of sphere) but at point P: 1)distance of point P from charge q is $\sqrt{17R^2}$ just use the formula of electric field for point charge then we get $$\frac{Kq}{17R^2}$$