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 Charge Density in a System of Concentric Spheres

This topic is largely a conceptual exercise. We will examine the different manifestations of charge density in a system of concentric conducting spheres as we ground, or do not ground, one of them.

Ungrounded Outer Shell

Figure this shows the setup for this topic. A solid conducting sphere of radius, Ra, is surrounded by a thick spherical shell of inner radius, Rb, and outer radius, Rc. We place a total charge, +Q, on the inner sphere. For now, the outer shell is not grounded. We will examine how the +Q charge distributes itself for this setup, and then consider how this changes when the outer shell is grounded.

Fig. 1: A solid conducting sphere is placed inside a thick spherical shell.

To begin, after the charge is placed on the inner sphere it will distribute itself across the outer surface. This is part of the definition of a conductor (and the beginning of the conceptual exercise here). The electric field is zero inside a conductor. Recalling Gauss' law,

we see that if a Gaussian surface (spherical) inside the sphere encloses zero charge, then the electric field will also be zero. If all of the charge on this inner sphere resides on the surface, then the surface charge density, σa, is,

and we already have enough information to determine the charge density of the surrounding shell.

The outer shell is also a conductor. The electric field inside this shell must also be zero. Gauss' law is used to figure out a way this condition can be met. Figure this exaggerates the thickness of the outer shell in order to better display a Gaussian surface residing within it. For any Gaussian spherical surface of radius, Rb < r < Rc, the electric field will be zero if the total enclosed charge is zero. That requires a total charge of -Q to reside on the shell. Again, however, since this shell is a conductor the charge must reside on the surfaces. It cannot reside on the outer surface at r = Rc because that would allow Gaussian spheres of radius, r < Rc, to enclose net charge.

Fig. 2: Distorted view of the system in which the thickness of the outer shell is expanded.

The only way for the electric field inside the shell to be zero is for a charge of -Q to reside on the inner surface. In such a case, the surface charge density of the inner surface of the shell, σb, is,

Since we did not place any free charge on the shell it must remain charge neutral. There can be no charge in the interior of the shell, yet the inner surface definitely has some charge, so this must be canceled out by an equal and opposite amount of charge on the outer surface. The charge density on the outer surface of the shell, σc, is,

Now that the total charge on the shell is returned to zero, the electric field outside of the entire system is determined solely by the inner sphere. Any Gaussian sphere that is placed with a radius, r > Rc, will enclose a total charge +Q. Outside of this shell, then, the electric field is equivalent to that of a point charge of +Q.

This can be taken a step further and the electric field for all space determined. Consider Gaussian spheres of various radii.

Since we have determined the electric field profile largely through conceptual considerations, let's solve for the potential profile to bring some mathematics to this topic. Knowing the electric field in space, the electric potential, V, can be calculated using,

where the integral is taken from the final position, f, for which the potential with respect to the initial position, i, is determined. Electric potential can only be calculated with respect to a defined reference. This is equivalent to gravitational potential energy in which the zero potential point is typically defined as the ground-level. In the case of electric potential we have the zero-point off at infinity. To solve for the electric potential at the origin we have,

where dl = dr r has been used because the only electric field is directed along the radial direction and the other coordinates are not necessary.

A separate integral is needed for each region of distinct electric field. Incorporating the electric fields that have already been determined gives,

and it is seen that the potential at r=0 is the same as the potential on the surface of the inner sphere (the first integral in this is zero). Another property of conductors is that they have a uniform electric potential. This is a requirement since the electric field is zero within a conductor, which means that the interior cannot add or subtract to the potential achieved by the surface.

Grounded Outer Shell

Now consider the outer shell is grounded. Specifically, the outer surface of the shell is grounded, as seen in figure this.

Fig. 3: The outer surface of the shell is grounded.

Grounding the outer surface means we force its potential to be zero. This must change the charge densities, electric fields, and potentials of the system. These changes can be worked out conceptually.

To begin, the charge density of the inner sphere cannot change. It is still a conductor with free charge placed on it. This charge must distribute itself evenly across the surface.

The surface charge on the inner surface of the shell must be the same as in the initial configuration as well. The shell is still a conductor and the only way to maintain a zero electric field within its interior is for a total charge that cancels that of the sphere to reside on its inner surface. This means the electric fields inside the sphere and in the region between the sphere and shell are unchanged from the previous, ungrounded, setup.

As to the charge density on the outer surface of the shell, though the potential there is zero the net charge cannot also be zero. No free charge has been placed on the shell and it must remain charge neutral. A charge canceling that found on the inner surface resides on the outer surface. Since the potential of the outer surface is zero that means the electric field between it and a position at infinity is also zero, once again shown in [expression on previous page].

The electric field outside of the outer shell (i.e., for r > Rc) is certainly zero if there is no charge on the outer surface. This is the case because any Gaussian spherical surface beyond r = Rb encloses no charge. The charges on the inner sphere and the inner surface of the shell cancel each other out. That leaves us to wonder how we will reconcile the requirement for charge on the outer surface with the additional requirement that the electric field outside of the shell be zero.

A final conceptual idea will solve this problem. Connecting a conductor to ground is equivalent to giving the conductor an infinite area. Remember this by considering that ground connections are often made by digging deep into the Earth (the absolute ground, in many ways). The Earth provides essentially infinite volume in which to dump free charge. If the outer surface of the shell now has effectively infinite area, then it can hold a charge that cancels out the inner surface while at the same time not generating an electric field. With a total charge +Q spread over an infinite area the charge density is zero.

The potential at the origin of this system is only slightly changed. According to [integral on previous page] , only the final integral has changed. The other electric fields have not changed from the ungrounded case. The fourth integral is now zero, which can easily by removed from [previously stated integral] to give a new potential value of,

and this concludes the review of the properties of this grounded system, all done based on conceptual understanding of the ungrounded system.

Initial Posting: Thursday, 31 January 2008
Last Updated: Thursday, 16 September 2010