This video is about field strength, potential and potential gradient.
We'll start by learning these concepts for an electric field between two plates,
then we'll move on to the electric field due to a point charge,
and since the concepts that we'll learn up to this point are very similar for gravitational fields,
we'll finish up by extending these concepts for gravitational fields.
Let's begin with the electric field between two parallel plates.
The electric field between two parallel plates is assumed to be uniform,
and it can be drawn like this.
The horizontal blue arrows are the electric field lines,
and the vertical black lines are equipotentials.
Here we assume the negative plates to be at a zero potential,
and the positive plate at a potential of 5 volts.
Let's recall two equations from subtopic 5.1.
The first one is for electric field strength,
so E is equal to F over Q, where E is electric field strength,
F is the force exerted on a charge placed in the electric field,
and Q is the magnitude of this charge.
The other equation is for potential difference.
Here, V is potential difference, W is the work done, when charge moves between two points in
the field, and Q is the magnitude of this charge.
Next, let's add another equation from subtopic 2.3.
This equation is work done by a force when moving an object.
W is work done, F is the force that acts on the object that is moving,
S is the displacement of the object, and theta is the angle between the direction of the force
and the direction of the displacement.
Let's assume that a positive charge with a charge of Q moves from the positive to the negative plate.
Rearranging the first equation from subtopic 5.1, we get that F is equal to E times Q.
Working with the equation from subtopic 2.3, and denoting the distance between the plates as S,
we get that the work done as the charge moves from the positive to the negative plate,
is equal to F times S. F is the force exerted by the field on the charge,
and since the field is uniform, this force is constant.
Since the direction of the field, hence the direction of the force,
is the same as the direction of the displacement, theta is equal to 0,
and since cosine 0 is equal to 1, we can simply omit cosine theta from this equation.
Replacing F by E times Q, we get that the work done is equal to E times Q times S.
Next, let's rearrange this second equation from subtopic 5.1 to give us W equals 2 V times Q.
Since both EQS and VQ are equal to the work done, we can write that EQS is equal to V times Q.
Canceling Q and rearranging gives us that E is equal to V over S.
This equation shows us that when the electric field is uniform,
electric field strength can be calculated as change in potential divided by distance moved.
Let's move on to draw the electric potential versus distance graph between the two plates.
We'll denote the distance with the letter R and the electric potential as VE.
We'll take both R and VE to be 0 on the negative plate.
We'll also assume that the distance between the plates is 5 cm.
Therefore, the equipotentials that you see on the diagram
have a distance of 1 cm between them.
Let's plot the graph.
On the negative plate, where R is 0, VE is also 0.
At a distance of 1 cm, the electric potential is 1 volt.
At 2 cm, it's 2 volts, 3 cm, 3 volts, 4 cm, 4 volts and 5 cm, 5 volts.
The last dot represents the positive plate.
Connecting the dots, we get a straight line.
Let's examine the gradient of this line.
Since we have VE on the vertical and R on the horizontal axis,
the gradient is equal to change in VE, so delta VE, divided by change in R, so delta R.
Let's link this gradient to the electric field strength equation.
Since we have a straight line, we can simply calculate the gradient
by picking the first and the last point on the graph.
These two points represent the two plates.
So in this case, delta VE is equal to the potential difference between the plates,
so V, and delta R is equal to the distance between the plates, so S.
Therefore, we can say that delta VE over delta R is equal to the electric field strength.
There is a final small change that we need to introduce here to be fully correct,
and this is adding a negative sign to the equation, giving us that E is equal to
negative delta VE divided by delta R. The reason for the negative sign is that the
direction of the electric field is opposite to how the potential of a positive charge varies.
In other words, as the positive charge is moved opposite the electric field,
it gains potential energy, and as it moves in the direction of the electric field,
it loses potential energy. Of course, a similar reasoning can also be applied to a negative charge,
but we won't go into these details here.
Delta VE over delta R is called a potential gradient, therefore the electric field strength
is equal to the negative of the potential gradient.
The formula that shows this connection between electric field strength and the potential gradient
can be found in the IB Physics data booklet, and here are the variables that are present in this
formula. Next, let's see how this connection works when the electric field is not uniform.
A non-uniform electric field that we learned about is one that arises due to a point charge.
We learned about the electric field due to a point charge in subtopic 5.1.
I won't repeat the related concepts here again, so feel free to go back to subtopic 5.1
and revise what we discussed. In this subtopic, we also introduced the following two equations.
Then, we combine them to get that E is equal to K times Q over R squared.
This is the equation for the electric field strength
due to a point charge Q at a distance R from the point charge.
Earlier in this video, we learned that electric field strength is equal to the negative of the
potential gradient. The discussion that follows from here will help us derive the equation for
electric potential at a distance R from a point charge. We'll use calculus in this derivation,
so don't worry if you find it a bit too complex. You do not need to know this derivation for your
IB physics exam. So using calculus, we can write the second equation for the electric field strength
as E is equal to negative dVe over dr. If we equate our two expressions for electric field strength,
we get a differential equation. We can solve this equation by the separation of variables technique,
which I won't discuss in detail here, but which gives us that Ve is equal to K times Q over R.
This is the equation in the IB physics data booklet for electric potential due to a point
charge, and here are the variables that are present in this formula. Before moving on to
gravitational fields, let's compare the electric potential versus distance and the electric field
strength versus distance graphs for a point charge. You can see that the general shape of the
graphs is similar, but since the potential graph is a 1 over R graph, while the electric field
strength graph is a 1 over R squared graph, there is a difference in the steepness of the graphs.
Next, let's move on to gravitational fields. The underlying concepts of different fields in
physics are very similar. What we learned about electric fields so far in this video
can also be applied to gravitational fields. Gravitational field strength is equal to the
negative of the potential gradient, which we can express using this formula. Here are the variables
that are present in this formula. Delta Vg over delta R is called the potential gradient.
Combining this formula and the one that we learned in subtopics 6.2 for gravitational field strength
and using calculus, we can show that Vg is equal to negative g times m over R. Here are the variables
in this formula. The derivation of these two equations is very similar to what we have seen
when we derived similar equations for electric fields. Therefore here we won't go into the details
of where these formulae come from. Just note that gravitational potential Vg is always negative.
We discussed the reasons for this in subtopic 10.1. Just as we have done for electric fields,
let's finish up with two graphs. One is gravitational potential versus distance and the other
is gravitational field strength versus distance for a point mass. Just as for electric fields,
you can see here that the shape of the graphs is similar and that the main difference is between
their steepness. Another important feature that we already mentioned before is that as you can see
gravitational potential is always negative. Let's summarize what we have learned in this video.
We discussed several concepts, but the main takeaway from here
are the four formulae that we discovered. Two of these are related to electric fields
and two are related to gravitational fields. For electric fields, we'll learn the connection between
field strength and potential gradient and the formula for electric potential due to a point
charge. For gravitational fields, we connected gravitational field strength with the potential
gradient and then saw the formula for gravitational potential due to a point mass.
As a final reminder, let me add that this expression and this expression are called
potential gradient. This completes our discussion of field strength, potential
and potential gradient. In the next video, we'll learn about potential energy.