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?

 A: 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 $\frac{kq}{r^2}=\frac{kq}{16R^2}$
A: 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. 
A: 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 
$$\frac{Kq}{17R^2}$$
