# Finding Intensity from an Electric Dipole

1. Apr 15, 2013

### RLOrion

The problem:

You and your engineering crew are in charge of setting up a wireless telephone network for a village in a mountainous region. The transmitting antenna of one station is an electric dipole antenna located atop a mountain 2.00 km above sea level. There is a nearby mountain that is 7 km away and is also 2.00 km above sea level. At that location, one member of the crew measures the intensity of the signal to be 4E-12 W/m2. What should be the intensity of the signal at the village that is located at sea level and 1.50 km from the transmitter?

Relevant Equations:

I=P/(4πr2)

Attempt at a Solution:

I decided to use the intensity measured on the opposite mountain to calculate the power of the antenna, and then use that answer to find the sea level intensity.

4*10-12=P/(4π70002)

P=0.00246 W

For sea level intensity:

r2=20002+15002 by Pythagorean theorem

I=0.00246/(4π(20002+15002))

Using this method, I got 3.136E-11 W/m2, which is apparently wrong. That's the only way I can think of to do the problem, so I'm stuck.

2. Apr 15, 2013

### TSny

Hello, RLOrion. Welcompe to PF!

Look at the note above problem 24 here .

3. Apr 16, 2013

### RLOrion

Thanks! I assume you're talking about the intensity varying by sin2(θ)/r2 if it's a dipole. Thanks for pointing that out; this is a Webassign problem, so that bit wasn't given.

I tried introducing this into the equation, but I still must be doing it wrong. It seemed to me that the power calculation wouldn't change as long as it's measured at the same altitude, assuming the dipole is vertically oriented. That would just make θ=90, so sin2(θ)=1.

I changed the expression for I at sea level to P*sin2(θ)/(4πr2)
I found θ by calculating tan-1(1.5/2), which equals 36.87 degrees. Sin2(36.87)=0.36, and so I multiplied my original answer by 0.36 to get 1.129E-11 W/m2, which was still incorrect.

Sorry, could you give me another hint?

4. Apr 16, 2013

### TSny

For a dipole antenna, the formula for the intensity would not be I = Psin2θ/(4$\pi$r2). Some of the constants in the formula would be different.

But all you need to know is that I is proportional to sin2θ/r2. Consider the ratio I2/I1 at the two locations.

EDIT: After plugging in numbers I get the same answer of 1.13 x 10-11 W/m2. I don't see where the mistake can be. I will think some more about it. One thing that sort of bothers me is that the problem isn't very clear about the exact location of the transmitter. At first I thought it would be up by the antenna, but that would put it more than 1.5 km from sea level. So, then I thought the transmitter might be at the base of the the mountain. But this apparently gives the answer we are getting. I noticed that all the numbers in the problem are the same as in problem 26 in the link I gave except for the distance between the two mountains which for your problem is 7 km. Is that right?

Last edited: Apr 16, 2013
5. Apr 16, 2013

### RLOrion

Yes, all the numbers except the distance are the same in the problem; Webassign changes some of the values for each question. I finally got the answer 55.7 pW/m2 using Webassign's practice questions, though I'm still not sure how to do the problem. I just divided the answer I was getting by the given answer in the practice problems, and found that my answer was off by the same factor each time.

What confused me was that the answer was greater than the one I was getting before I took θ into account. This seemed to imply that the angle of the antenna couldn't be vertical, and was closer to perpendicular with the ground receiver than the one on the opposite mountain. If that's the case, I wouldn't know how to even measure the power.

6. Apr 16, 2013

### TSny

I ran across someone's solution to the same problem with all the same numbers except for the distance between the mountains (which they have as 4.0 km).

They take r to be the distance between the transmitter and the village (1.5 km ). I don't see how this can be if the antenna is 2.0 km above sea level and the village is at sea level! I keep trying to visualize how that can be.

Also, their solution uses θ = 53.1o (the complement of 36.9o). Again, I can't understand why. I keep thinking that I'm misinterpreting the wording of the problem.

Last edited: Apr 17, 2013
7. Apr 17, 2013

### RLOrion

Yeah, I don't see why r would be that distance either. It seems to me that 53.1° would be correct if the angle of the antenna was horizontal, but that would cause the intensity at the opposite mountain to be measured as zero.

I just did the math using the method you described and got the right answer! I'm surprised that the 2 km height wasn't relevant, however.

Last edited: Apr 17, 2013