A hollow insulating sphere with inner radius, R1, and outer radius, R2, contains

Question

A hollow insulating sphere with inner radius, R1, and outer radius, R2, contains charge described by a charge density (C/m^3), , given by = 0 for 0 <= r <= R1; = D/r^2 (C/m^3) for R1<= r <= R2, where D is a positive constant with units of (C/m); and = 0 for r > R2.

1. Draw a picture showing the charge distribution and the resulting electric field line pattern. If you wish to use separate diagrams to show the charge distribution and the electric field, that is fine. Make sure that you show the electric field lines in all regions of space including all regions where there is no charge and also the regions   where   the   charge   resides.       To   draw   the   electric   field   pattern   use superposition, that is, the notion that the net electric field at any point in space is the sum of all the electric field contributions from each small part of the charge distribution. The charge distribution can be divided any way that you like as long as you know the field pattern resulting from each part. Use symmetry arguments to aid in this endeavor if at all possible.   Your drawings need to be accompanied by some written explanations of the reasoning allowing you to make the drawings.
2. If possible, pick a closed “Gaussian” surface or surfaces appropriate for the geometry of the charge distribution at hand such that the electric field can be determined.
3. For appropriate cases of charge distributions explain (combinations of equations liberally sprinkled with words of explanation are appropriate here) how Gauss's law can be applied to find the electric field in all regions of space. Then demonstrate the technique by finding mathematical expressions for E in those regions.
4. If Gauss's law cannot be solved for the electric field, explain clearly and specifically why not.

Explanation / Answer

At all points inside the charged shell the electric field E=0as the charge inside is 0.

We can construct a Gausian surface of radious r >R2, shown as dashed line. At all points of this sphere the magnitude os the electric field is same and the direction is perpendicular to the surface.

                                E =

Electric field at any external point is same as if the total charge is concentrated at the center.

Thus we can write

                  Einside =0

                    Eoutside =

                    Esurface   =

At all points inside the charged shell the electric field E=0as the charge inside is 0.

We can construct a Gausian surface of radious r >R2, shown as dashed line. At all points of this sphere the magnitude os the electric field is same and the direction is perpendicular to the surface.

                                E =

Electric field at any external point is same as if the total charge is concentrated at the center.

Thus we can write

                  Einside =0

                    Eoutside =

                    Esurface   =

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