5.6 Spherical Capacitor from Office of Academic Technologies on Vimeo.

**5.06 Spherical Capacitor**

A spherical capacitor consists of two concentric spherical conducting plates. Let’s say this represents the outer spherical surface, or spherical conducting plate, and this one represents the inner spherical surface. Let us again charge these surfaces such that by connecting the inner surface to the positive terminal of the power supply of a battery and the outer surface to the negative terminal of the power supply. In doing that, the inner surface will be charged positively to some plus q Coulombs and the outer surface will be charged negatively to some minus q Coulombs. Then give us some dimensions, say inner radius is a and outer radius is b. Therefore by charging the capacitor, we completed the first step to calculate the capacitance of this spherical capacitor.

In the second step, we’re going to calculate the electric field between the plates; therefore we choose an arbitrary point between the plates. Since the geometry is spherical geometry, then we will choose a spherical Gaussian surface such that it is passing through the point of interest. Let’s say that point is r distance away from the center of the spherical capacitor. Gauss’s law says that integral of E dot dA over this closed surface s is equal to q enclosed over Epsilon zero, which is the net charge inside of the region surrounded by Gaussian surface.

Well, the electric field, after we charge these plates, is going to be originating from the positively charged plate and entering into the negatively charged plate, therefore it is going to be in radially outward direction, filling the region between these two charged plates. Therefore the electric field is going to be pointing radially outward from positive plate to negative plate. The surface area vector, which is incremental surface area vector, is perpendicular to that incremental surface on this Gaussian sphere. Therefore that, too, will be in radially outward direction and perpendicular to the surface like this so the angle between E and dA at every point along this surface will be zero degrees.

E dA cosine of zero will be the explicit form of this dot product integrated over this closed surface is equal to q enclosed over Epsilon zero. Cosine of zero is 1, and as long as we are on the surface of this Gaussian sphere, we will be the same distance away from the source charge that it encloses. So this will be constant over that surface. We can take it outside of the integral, which will therefore give us E times the integral of dA over this Gaussian sphere.

That will be equal to q enclosed over Epsilon zero. Integral of dA, again, by adding all these incremental surfaces along this Gaussian spherical surface will eventually give us the surface area of that sphere, which is 4 Pi times the radius squared, r squared. The net charge enclosed inside of the region surrounded by this sphere is the total charge distributed along this inner sphere because it encloses that sphere completely. That will be, therefore, the amount of charge along that inner sphere, which is q. We will have q over Epsilon zero.

Solving for the electric field, from here between the plates, will turn out to be q over 4 Pi Epsilon zero r squared. If you look at this expression, we see that it is identical to the point charge electric field. And as you remember, we mentioned that a spherical charge distribution, either solid sphere or spherical shell, behaves like a point charge for all the exterior regions. So this point p, our point of interest over here, is the exterior point for this spherical distribution here. Therefore it is behaving like a point charge, as if it’s all charge q concentrated at its center so that we end up with electric field is equal to q over 4 Pi Epsilon zero r squared.

Now, once we determine the electric field magnitude, which is in radially outward direction, we can go ahead and calculate or apply the third step, which is the potential difference between the plates is equal to integral from positive to negative plate of E dot dl. Again, like in the case of cylindrical capacitor, we’re going to choose a path in this case, in order to calculate this path integral, in radial direction. And since our electric field is radially outward, we will choose the path to take the integral also in radial direction, and therefore dl is going to be equal to dr for this radially out path.

In other words, the incremental displacement vector dl will be replaced by the incremental displacement vector in radial direction. Then the potential difference between the plates becomes E magnitude, which is q over 4 Pi Epsilon zero r squared. dl magnitude will be dr magnitude times cosine of the angle between these two vectors.

Again, we are choosing this path such that it coincides with the electric field vector so that the angle between these two vectors will become zero degree. Cosine of zero is just 1. Here q and 4 Pi Epsilon zero, these are constant, so we can take it outside of the integral, and that leaves us the potential difference between the plates of this capacitor as q over 4 Pi Epsilon zero integral of dr over r squared. Now, the integral is taken from positive to negative plate. For that region, if we look at the change in the radial direction, we will start there from inner radius up to the outer radius, so boundaries of the integral will go from a to b.

Moving on, potential difference will be equal to q over 4 Pi Epsilon zero integral of dr over squared is minus 1 over r, which will be evaluated at a and b. Substituting the boundaries, potential difference will be equal to q over 4 Pi Epsilon zero, and we will have minus 1 over b minus, and another minus will come from the function, will make this 1 positive over a. If you have a common denominator over here, we can write down this expression as q over 4 Pi Epsilon zero b minus a over ab. So that’s the potential difference between the plates of the spherical capacitor.

The last step says that calculate the capacitance from its definition, which is the ratio of the magnitude of charge in the capacitor to the potential difference between the plates of the capacitor, which will then be equal to–for the potential difference–we have q over 4 Pi Epsilon zero, times b minus a over ab.

The charges will cancel, and the capacitance of a spherical capacitor will turn out to be 4 Pi Epsilon zero times ab over b minus a. Like in the previous cases, for the parallel capacitor and cylindrical capacitor, here again we see one more time that the capacitance is directly dependent to the physical properties of the capacitor. In this case, for the spherical capacitor, inner and outer radius of the capacitor.

Here we are going to consider a special case. If you consider a charged cylinder, we can always treat that cylinder like a capacitor such that its outer plate located at infinity. In other words, we can say that let a is equal to R and b goes to infinity. So we take the outer sphere to infinity, and we’re talking about a charged, isolated sphere. This case corresponds to a charged isolated sphere. Then we can express this expression, capacitance, as 4 Pi Epsilon zero. And let’s say replace a with big R and leave b as b for a moment and write down the denominator in b parentheses. So take b outside, 1 minus, replace a with R, R over b.

The “b”s in the numerator and denominator will cancel, and as b goes to infinity, then R over b will approach to zero, so the capacitance of an isolated charged sphere will turn out to be 4 Pi Epsilon zero, taking R over b to zero. We will end up just with r, indicating that the capacitance of an insulated, charged sphere is equal to 4 Pi Epsilon zero times the radius of that sphere.