APPLICATIONS OF GAUSS S LAW Although Gauss s Law is always correct it is generally only useful in cases with strong symmetries. The basic problem is that it gives the integral of E rather than E itself. To get E we need to be able to take it outside the integral, and this is normally only possible when we have symmetry. We consider three such cases which are especially important. SPHERICAL SYMMETRY Consider the system shown, consisting of two hollow, concentric, spherical, conducting shells. A charge Q is placed at the center, a charge Q on the inner shell, and a charge Q on the outer. We wish to find the electric field everywhere and the charge distribution on the shells. First we need to talk about conductors. CONDUCTORS A conductor is a material in which charges are free to move macroscopic distances. By this we mean that they are free to roam about the material and are not restricted to a particular atom. In any material the charges can move in their atom, but in conductors they can leave the atom. An immediate consequence is that in the presence of an electric field they will move because of the force it exerts on them. They will continue to do so until the field is zero. Sometimes, as in electric circuits, this cannot be achieved and the motion continues indefinitely. However for the present we will be interested in the case in which there is not a complete circuit, so that a zero field condition can be attained. This situation is called electrostatics. In the static case we immediately see that the field everywhere inside the conductor must be zero. Next consider the surface of the material. Here the component of the field along the surface must be zero (otherwise charges would move along the surface until it was). There can be a field outside the conductor, but it must be perpendicular to the surface. Finally we note that since the field is zero everywhere inside the material we can draw an tiny sphere around any point in the interior
and the integral of E over the sphere will be zero. This means that there is no net charge inside the sphere. Since we can do this at any point we conclude that the charge density inside the sphere is zero everywhere. How can this be, since we know that the material is made of atoms which are protons, neutrons, and electrons? The answer is that we must draw the sphere large enough to include atoms this simply acknowledges the fact that the charges can rearrange themselves inside the atom or molecule. Hence we find three critical properties of conductors in electrostatics: the electric field is zero everywhere inside the conductor; the field outside the conductor must be perpendicular to the surface; the net charge at any point in the interior of the conductor must be zero. We now return to our spherical problem. SOLUTION OF SPHERICAL PROBLEM Because of the symmetry we know that the field must be everywhere radial and the same in all directions. In other words it can depend only on the distance from the center and can have only an r component. We begin by drawing an imaginary sphere centered at the origin and of radius r where r < R. We now evaluate EdA We note that the perpendicular to the sphere is the radius. Hence da points in the ˆr direction. But so does E (could be in or out, but is parallel to ˆr ). Thus: EdA EdA But we also know that E is the same in all directions. Thus it is constant over the imaginary sphere and we can take it out of the integral: But the area of a sphere is 4πr. Hence: EdA EdAEdA EdA E4r But the total charge inside the imaginary sphere is just Q. Hence: EdA E4r kq E r o rˆ
The amazing thing about this result is that the charges outside r have no effect on the field inside r. This depended on the symmetry and we will need to examine it more closely in the general case. Next we draw an imaginary sphere inside the inner shell. Since the field is zero inside the conductor we find: EdA 4kQ Q IN IN Thus the total charge inside r must be zero. This could consist of three parts: Q, the charge on the inner surface of the inner shell, and the charge inside the shell between its inner surface and r. We know that the latter is zero. Hence we must have: Q Q Q Q inner inner Thus the charge on the inner surface of the inner shell is Q. But the total charge on the inner shell is Q. Thus, since there is no charge in the conductor we must have a charge Q + Q on the outer surface of the inner shell. Next draw an imaginary sphere between the two shells. The situation is now exactly the same as in the case r < R except that the charge inside is now Q + Q. Proceeding as before we can solve the entire problem, finding: kq r R E rˆ r RrR E k Q Q R r R 3 E rˆ r R3 rr4 E r R 4 k Q Q E r r ˆ Q inner Q Q Q Q outer Q Q Q inner Q Q Q Q outer
CYLINDRICAL SYMMETRY We next consider the case of two hollow, coaxial, conducting cylinders with a line charge on the axis. The inner cylinder has a total charge Q, the outer has Q and the line charge is Q. The length of the cylinders is L. This time the symmetry is more complicated. Obviously the field is the same in all directions about the axis, but it need not be the same everywhere along the axis. Hence if we use cylindrical coordinates with z along the axis we have: E E,z ˆ E z,z zˆ Now suppose we were to go a long ways away from the cylinders. Then the arrangement would look like a point charge Q + Q + Q. We know what the field would then look like: But we also know that it must start perpendicular to the conductors. Hence it must be like:
Thus if we stay away from the ends the field will be only in the ˆ direction: E Ep,zˆ The next question is what the z dependence of the charge is. Since like charges repel, we would expect the charge density at the ends of the cylinders to be larger than at the middle. This is because at the middle there are equal charges on each side pushing in opposite directions, while at the ends there are only charges on one side. In the case of a flat disk for example, the charge density is twice as big at the edge as at the center. However if we stay away from the ends we expect the density to be pretty uniform. Another way of looking at it is to note that far from the ends we cannot see either end and hence neither direction is special. This means the charge density should not look different in one direction as opposed the other. That means it must be uniform. We will assume this to be the case. Then: E E( )ˆ We can now solve the problem as before by using Gauss s Law. We draw an imaginary cylinder of radius r where r < R. Then: EdA Er Q Q L IN Er E 4 kq L kq Lr We now draw the imaginary cylinder inside the inner cylinder. As before the field is zero and hence the charge inside must be zero. This means there must be a charge Q on the inner surface of the inner cylinder. Since there is no charge in the conductor this requires a charge Q + Q on the outer surface of the inner cylinder. ˆ
Proceeding in the same fashion as for the spheres we find: kq ˆ r R E Lr RrR E k Q Q R r R 3 E ˆ Lr k Q QQ r R 4 E ˆ Lr Q Q inner Q Q Q outer Q Q Q inner Q Q Q Q outer Notice that the field now falls off as /r rather than /r. This is because the field lines now spread out around a circle rather than over a sphere. Of course if we go far from the cylinders the approximations break down and we revert to /r. PLANES We know consider the field due to a charge on a conducting plane. We do it in the same fashion as for cylinders. As long as we stay away from the edges and close to the planes (distance from plane << dimensions of the plane) we expect the field to be perpendicular to the plane and uniform: We draw an imaginary rectangular box as shown:
Then by symmetry we expect the field to be the same on both sides of the plane. Since it is perpendicular to the plane, no lines go through the sides, and we get: EdA Ea Q IN a Q A a kq Q Ea 4 kq E A A g Note that this time the field does not fall off with distance. Of course it doesn t since the lines are all perpendicular to the plane and hence parallel to each other. As before this is only true close to the plane. FINDING THE CHARGE FROM THE FIELD We now consider the inverse problem of finding the charge inside a volume given the field at the surface of the volume. Contrary to the problems we have just done, this can always be solved. As an example consider the volume enclosed by the surface formed by isosceles triangles: located with vertices at (,,), (,,), (,,) and (,,), (,,), (,,) plus the necessary connecting planes. Hence the volume is half a cube of side, with one corner at the origin and lying in the first octant. Suppose the field in the region of the object is: E E xyxyzyzxz ˆ ˆ ˆ
We need to evaluate the integral over five faces. First consider the triangle in the xy plane (z = ): E E xy xˆ da dxdyzˆ E E da xy dxdy xˆz ˆ Hence EdA face Next do the integral over the triangle at z =. x y E xyxyyxz E ˆ ˆ ˆ da dxdy zˆ E EdA x y x x E E E E da x x dxdy x xy dx x x x dx 3 x x 3 3 6 E E E
Next do the integral over the bottom face: E E zx zˆ da dx dz yˆ EdA EdA Next the integral over the face in the yz plane: E E yz yˆ da dydz xˆ EdA EdA Finally we need the integral over the face containing the hypotenuse of the triangles. This is a bit more complicated because da has two components. E xyxyzyzxz E ˆ ˆ ˆ da da xˆ yˆ E da EdA xyyz da dz dg where
/ dg dx dy But yxdy dx / dg dx dx dx da dxdz Thus E E da xy yz dx dz x(x) (x)z dx dz Hence E z x z x 3 E E z z E z z E 6 4 6 4 total x x x z x dz z dz 3 6 E 7 E Ed 6 6 4 Then from Gauss s Law we get: 7 E 7 E 48k 4kQ IN QIN