# Homework Help: Few electromagnetics questions

1. Nov 1, 2006

### Cairrd

Just doing an exercise and I am stuck on the following questions. Any help would be appreciated:

1) A vertical dipole, with charge +40C at 10km high, and -40C at 6km high, what is the E field at the ground directly below (ground assumed perfect conductor). And at what distance on the ground will the E field be 0?

2) A coaxial cable is constructed using 2 dielectrics. Dielec 1 occupies 1/6th of the total area between the conductors who's radii are 1 and 7mm. Relatice Perm. of E1 = 5, E2 = 3. Calculate capacitance per unit length.

3) A small dipole centered at the origin, has components (0,0,p). By differentiating the expression for potential at (x,y,z) find the electric field components of x, y and z for (x^2 + y^2 + z^2) >> d^2

where p = dq, q is the charge and a is the seperation

Think I have the others sorted, just these left. Thanks, David.

2. Nov 1, 2006

### Staff: Mentor

You need to show some of your own work on these in order for us to help you. What have you done on -1- so far?

3. Nov 1, 2006

### quasar987

You can do the first part of a), right?

The second part sounds more difficult. You need to know how the conductor below react (i.e. how charges will arrange themselves on the surface). But this is a classic problem used to illustrate the mthod of images. If you've seen the method of images, you know how to do this.

4. Nov 1, 2006

### Cairrd

When I have used the Method of Images it has only ever been with 1 point charge, how does it work with a dipole?

5. Nov 1, 2006

### quasar987

Good question. Would the effects simply be additive? Intuitively, it seems reasonable, but how to justify it?

In the case of a single point charge, the induced surface charge density at the surface of the conductor is a result of the field of that charge. If you have two charges, the resulting field is the sum of the individual fields, hence the resulting induced charge is the sum of the individual ones.

Or you can see it this way. First there was the void. Then bring the ground from infinity to its final position. Then bring the first charge from infinity to its final position. This induces a surface charge that maintains the potential at 0 at the conductor. And you know how to calculate that. Then bring the second charge from infinity to its final position. This also induces a surface charge on the conductor in order to keep it at 0 pot. The net surface charge is the sum of the two (one is a positive density, the other is a negative, but the resulting will be positive become the negative charge in the dipole is closer to the ground)