Capacitors

What is a capacitor? Basically, it is any two conductors separated by an insulator. In practical applications, the conductors usually have charges of equal magnitude and opposite sign, so the net charge of the capacitor is zero. The electric field between the two conductors is proportional to the magnitude of this charge, and the potential difference V_ab between the conductors is also proportional to the charge magnitude Q.

Electric Potential

When charged particles move in an electric field, work is being done by the field on the particles. In this special case the work can be expressed in terms of potential energy which is associated with something called electric potential. It is the potential that the particle has to do work, and since it is in an electric field, electric potential seems like the most logical name for it.

Application’s of Gauss’s Law

Location of Excess Charge On a Conductor

We know and will prove that the electric field E = 0 at all points within a conductor when the charges in the conductor are at rest. We can make an imaginary surface in the interior of a conductor, such as surface A in the illustration at right. Becuase E = 0 everywhere on this surface, the net charge inside the surface has to be zero. Now compress that surface so it encloses just a point, and since E = 0 everywhere on the surface, the net charge inside that point is zero. That also means that there cannot be a net charge at any point within the conductor, because that tiny point could be put anywhere in the conductor. If that’s so, that means all the charge must be on the outer surface of the conductor, as shown.

Gauss’s Law

Kal Friedrich Gauss (1777-1855) was a German scientist and mathematician known for a relation known as Gauss’s Law. His law was a statement of an important property of electrostatic fields. With your understanding of field lines, it will be easy for you to understand Gauss’s Law.

The field of an isolated electric charge q can be represented by field lines that radiate outward, right? Let’s say we surround this charge with a sphere of radius R with the charge being at the center. This sphere’s surface is, therefore 4pR². So if the number of field lines from the charge q is N, then the number of lines per unit surface area on the sphere is N/(4pR²).

Now what if we surround this sphere with another one, with a radius of 2R? The surface area is now 16pR² and the number of lines per unit area is only N/(16pR²). The line density is only a fourth of the first sphere, which makes sense since the radius of the second sphere is twice that of the first.

However, the total number of lines emanating through at the 2R sphere is the same as the R sphere. The field is inversely proportional to R squared, and the area of the sphere is proportional to R squared. So that means the product of the two is independent of R. Now for a sphere of a variable radius r, the magnitude E of the surface is:

The surface area of the sphere is:

A = 4pr²So if you multiply the two you get:

This means that the product of the two is independent of r and depends only on the charge of q (since we are given everything else).

Also, what is true of the entire sphere is true of any part of its surface. Thus an area DA is outlined on the surface of a sphere, seen in the illustration to the right (click on the illustration for a VRML model of the Gaussian projection), of radius R and then projected onto the sphere of radius 2R (same example as before). The area projected on the larger sphere is 4DA, which proves that the product EDAis independent of the radius of the sphere.

The reason for explaining this is for explanation of nonspherical surfaces. Let’s say we replace the second sphere with an irregular shape as seen in the illustation to the left (click on the illustration for a VRML model of the irregular shape). Think about the small area element DA. The area is larger than the area of a sphere if its distance from q were the same. Now if a normal to the that surface makes an angle q with a radial line from q, two sides of the area projected on the spherical surface are foreshortened by a factor cos q. Thus the quantity of EDA for the spherical surface is EDA cosq for the irregular surface.

To sum it all up, divide the entire irregular surface into small pieces of DA. Calculate EDA cos q for each piece, and then add them together. So this is what we get, as long as the surface encloses the charge q:

Thus, a surface that encloses no charge (q = 0) produces the following result:

Now if you realized that the field varies from one point to another on the irregular surface, you are correct. This means that Equation only applies when the limit of the area elements become really really small. This means that we must bring in integration, and in this case, the limit of the sum becomes a special surface integral of E cos q, written as follows:

You probably aren’t used to the integral with the little circle on it, but it just tells you that it is taken over a closed surface enclosing the charge q (called a surface integral). Now we also know that E cos q is the component of E perpendicular to the surface each of the points so we will use a special notation (with the perpendicular sign) in replacing E cos q, which will make the equation look like this:

This quantity is called the electric flux through the area dA. Electric flux is denoted by Y, and an element of flux of the small area dA is dY. Thus the following relationships are true (because we just defined them):

The total flux Y through a finite surface is the integral of dY, or:

Thus, we make the necessary substitutions to come to this brilliant conclusion:

Now what about multiple charges? We only talked about one charge, inside or outside, but only one charge. But remember that the total electric field E at a point on the surface is the vector sum of the field produced by any number of single charges within the surface. Which means that we could have one or one hundred charges within the surface, only the vector sum of these charges will be taken into effect and the vector sum acts as if it was just one charge. So to be thorough about the equation, it can be written as:

We called the sum of all the charges q_enc to stand for all the enclosed charges.

Here is a quick list of rules that will help you calculate electric flux:

• If E is at right angles to a surface of A at all possible points and E is constant at these points, then:
  
• If E is parallel to the surface at all points, the perpendicular component of E = 0 and so does the integral.
• If E = 0 at all points of a surface, the integral is zero.
• The surface you apply Gauss’s Law to doesn’t need to be a real physical surface. In most applications you use an imaginary surface that is actually empty space, embedded in a solid body, or partly in space and partly in body.
• In the integrals, E is always the total electric field at each point on the surface. In general, the field is caused partly by changes within the volume and partly by charges outside, but even when there is no charge in the volume the field on the surface isn’t necessarily zero. But in that case, however, EdA is always zero.
• In using Gauss’s law to calculate fields, you need to know which surface to choose. The point or points at which the field is to be determined must lie on the surface and that the surface must have enough symmetry so that you can evaluate the integral. So if the problem has spherical symmetry, the easiest Gaussian surface (as it is called) to use would be a sphere.

Electric Field

In this section, we will be working with electric fields and what they look like with different shaped conductors. To review, here are a few rules about electric field lines:

• No field lines originate or terminate in the space surrounding a charge. It has to start at a positive charge and end at a negative charge.
• Field lines never cross. At any one point, the resultant field (and the field lines that represent it) can have only one direction.
• Since the field exists at all points, you could technically fill all of space with field lines. But you can represent it with only a number of field lines that can be used to indicate the magnitude of the field and its direction. The closer the lines, the denser and stronger the field.

Also, we should review Coulomb’s law. If you recall, Coulomb’s law is used to calculate the magnitude of a force generated by an electric field. If we had a charge q, the force on charge q‘ that is a distance of r away from it is:

And the electric field at charge q‘ is:

Now that we reviewed these concepts, we can begin deriving. The next three derivations will give you some experience with the various types of concepts and integrals that we will be dealing with later.

---

Ring-Shaped Conductor

The conductor is a ring of radius a, and carries a total charge of Q. Now we are going to find the electric field at a point x from the center along a line perpendicular to the ring. You can see the illustration at right to visualize this setup (click on the illustration for a VRML model of the ring).

First you need to break down the ring into sections of size ds because you will need to integrate along the ring. The charge of the segment dr is dQ, a small part of the total charge. At point we are looking at, the element of charge causes an electric field contribution dE having a magnitude of:

Now just look at the figure. If we were to break down the components of the electric field contribution of dQ into x and y components, what do you think would happen to the y components of the field? It turns out that they will cancel out with the y components on the opposite side of the ring. This means that we only need to consider the components of dE_x of this field, and it is defined as follows:

So to find the total of all the x components, simply integrate:

On the right side everything except dQ is a constant so they can be moved outside the integral. Then integrate to get:

That is your answer folks.

---

Long Charged Wire

Now you have a long charged wire of length 2L that lies along the y-axis and has a charge per unit length of l. We will find the electric field caused by this wire at point P, a distance r from the midpoint of the wire. You can see the illustration at right to visualize this setup (click on the illustration for a VRML model of the wire).

Now, just like before, divide the wire into infinitesimal segments dy, and the charge of which is known to be dq, where dq = l dy. The distance s from any segment at position y to point P is:

Thus, the contribution dE to the field at P due to this piece is given as:

To find the total field at P we need to sum the contributions of all the segments, which means we need to integrate. However, we need to break it all down to components first. Again, because of the symmetry, the y components are going to cancel out, and we only need to worry about the x components:

The integraton can be done by trigonometric substitution or by a table of integrals. Since this isn’t calculus I trust that you will go look on your own. Once you integrate it you get:

Now, if the length of 2L is much longer than the distance r (if the wire is basically infinitely long), then r becomes negligible compared to L inside that square root. So effectively it simplifies down to:

Jackpot! You are done with this incessantly long thin wire.

---

Infinite Plane Sheet of Charge

This is the last one, we promise. A positive charge is distributed uniformly over an entire xy-plane, with a surface density of charge s. We will find the electric field at point P, a distance a from the plane. You can see the illustration at right to visualize this setup (click on the illustration for a VRML model of the sheet).

Now here is where we “cheat” a little. Divide the sheet into long strips of width dx, parallel to the y-axis. Now do you notice how they resemble the long charged wire? We can use that to help us along.

But first the area of a portion of the strip of length L is Ldx, and the charge dq on the strip is:

dq = sLdx
So the charge per unit length is:

With the equation from the long wire we now setup dE in the xz-plane now:

Now again we use symmetry and we notice that all the contributions in the x direction cancel each other out and only the upward field in the z-direction matters. So the resultant electric field at P is in the z-direction and can be defined as:

dE_z = dE sin q
Since there are dE‘s on both sides, naturally we think to integrate and we get the following:

Before you start going crazy wondering where a actually went, it was the distance between the point and the plane and one would imagine that value would be pretty important, but what it tells you is that the intensity of the field created by the sheet is independent of the distance from the charge. More simply, the field is uniform and normal to the plane and does not weaken when you get away from the sheet.

Angular Momentum

Angular momentum works almost like linear momentum. If an object is spinning, it is said to have some amount of momentum that is based on its rotational inertia and its angular velocity (similar to mass and linear velocity for linear momentum). Angular momentum (L) is defined as:

L = Iw
As you can see, it is very similar to the linear momementum equation, p = mv. For a spinning point particle, the angular momentum is:

L = mvr
Just like linear momentum, angular momentum is always conserved as well when there are no external forces acting on a system. This is why skaters spin faster when they bring their arms inward. Bringing their arms inward decreases their rotational inertia, and since angular momentum must be conserved, angular velocity increases.

In another case, if we have two spinning disks spinning at different rates and then join together to spin at the same speed, the angular momentum will still be constant. For example, disk A is spinning at 3 s^-1 and B is spinning at 5 s^-1. Disk A has a mass of 9 kg and has a radius of 0.30 m. Disk B has a mass of 4 kg and has a radius of 0.20 m. Let’s figure out what happens when they come together:

L_before = L_after I_Aw_A + I_Bw_B = (I_A + I_B)w_after
If you plug in all the values, you eventually will find out that the two disks joined together spin with an angular velocity (w_after) of about 3.3 s^-1.