Hi. It's Mr. Andersen and this is AP Physics essentials video 37. It is on the average
value of the electric field. And so we have used topographic maps and topographic profiles
as an analogy to understand how electric charge works in an electric field. And so one of
the places I always wanted to go to is Ayres Rock in Australia. This is a picture of it
from the side. And this is what it looks like from above. And this is going to show the
contour lines on the side. In other words lines of equal elevation. And so there is
a wonderful website you can go to, it is geocontext.org. And what you can do is take any map in the
whole planet and you can drag from A to B. So this where I have chosen, from point A
to point B. And then it will build the topographic profile on the side. And so what you can do
is you can actually move point B and it is going to show you a different topographic
profile. You can then zoom in and we can to do different areas on the side and we can
see how much it is changing. In this case it was 32 degrees. If we look at the whole,
going up this side it is going to be 22 degrees. It is showing how much it is changing in the
vertical in relation to the horizontal. This looks really steep, 57 degrees. Quite a change
right there. If we were to look on the top of Ayres Rock we could see that it looks more
flat along the top, even though the profile gets exaggerated, it is only 2.5 degrees,
the average change right there. And so again, if we are using this analogy and now applying
it to electric potential, if we have an electric field like this, between two parallel plates,
we see equipotential lines that are equally spaced out. And so you can think of this almost
like an inclined plane. It does not matter where you step in, it is going to be the same
electric field strength on either side. And again, to figure that out we take the voltage
across these two parallel plates divided by the displacement by the two and that is going
to be our electric field strength. And it is going to be consistent here. But what if
we do not have a consistent electric field? What if we have these equipotential lines
that are spaced out? Well then we can just use the average field strength. And so we
would find the potential at two different areas. In this case 12.6 and 2.8. And then
we are going to divide the difference between those two by the displacement between the
two. Now it is not exact, but it is going to give us an average electric field strength
or electric potential. And so again if you have parallel plates like this, we have got
the voltage and displacement. Voltage remember is a measure of how much work is done to move
a test charge from one place to another. So if we are figuring out the electric field
strength all we do is divide the voltage by the displacement. And so if the voltage is
1.5 and the distance between the two is 5.0 centimeters, we just use this equation. What
is going to be our change in voltage? It is going to be 1.5. What is our displacement?
You can see right here that I have changed this from 5.0 centimeters to 0.05 meters.
And now I simply multiply it and that is going to be my electric field strength or electric
potential between the two. But what if we are given a problem like this. This is really
common in AP Physics. How do I figure out the difference or the average distance in
those potentials, let's say between point A and point B. Well all you do is find what
the potential is at those two different areas. Just like finding the elevation at two different
areas on a topographic map. Right here it is 15 volts. Right here you can trace it.
Right here it is down 0 volts. And so we are going to have a change in potential of 15.
Where is my distance? Where is my displacement? Well you can see here on the y-axis that it
is showing my displacement in meters. And so I set up my equation again as a change
in the voltage or change in the potential over the change in displacement. So I am getting
15 volts as a difference between the two. And then how did I get 2.2 meters? Well this
look like around 6 and this looks like around maybe 3.8. I am not perfect obviously. But
that is going to be the distance between those two. And so then I would just say it is 6.8
volts per meter. Now a real common other question might be, let's say we put point A here and
we put point C right here on that line over here. What would be the difference or the
average value in the electric field? Or the difference? It is going to be 0. If you ever
move along a isoline, again it is going to be equal levels of potential. And so did you
learn to calculate the average value? Here is our equation. Again it is on the formula
sheet. And then did you learn to use isolines? You just trace the isoline. That is going
to give you its potential at different areas on the map. I hope that makes sense. And I
hope that was helpful.