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This video is about field strength, potential and potential gradient.

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We'll start by learning these concepts for an electric field between two plates,

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then we'll move on to the electric field due to a point charge,

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and since the concepts that we'll learn up to this point are very similar for gravitational fields,

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we'll finish up by extending these concepts for gravitational fields.

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Let's begin with the electric field between two parallel plates.

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The electric field between two parallel plates is assumed to be uniform,

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and it can be drawn like this.

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The horizontal blue arrows are the electric field lines,

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and the vertical black lines are equipotentials.

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Here we assume the negative plates to be at a zero potential,

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and the positive plate at a potential of 5 volts.

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Let's recall two equations from subtopic 5.1.

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The first one is for electric field strength,

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so E is equal to F over Q, where E is electric field strength,

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F is the force exerted on a charge placed in the electric field,

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and Q is the magnitude of this charge.

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The other equation is for potential difference.

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Here, V is potential difference, W is the work done, when charge moves between two points in

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the field, and Q is the magnitude of this charge.

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Next, let's add another equation from subtopic 2.3.

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This equation is work done by a force when moving an object.

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W is work done, F is the force that acts on the object that is moving,

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S is the displacement of the object, and theta is the angle between the direction of the force

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and the direction of the displacement.

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Let's assume that a positive charge with a charge of Q moves from the positive to the negative plate.

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Rearranging the first equation from subtopic 5.1, we get that F is equal to E times Q.

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Working with the equation from subtopic 2.3, and denoting the distance between the plates as S,

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we get that the work done as the charge moves from the positive to the negative plate,

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is equal to F times S. F is the force exerted by the field on the charge,

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and since the field is uniform, this force is constant.

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Since the direction of the field, hence the direction of the force,

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is the same as the direction of the displacement, theta is equal to 0,

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and since cosine 0 is equal to 1, we can simply omit cosine theta from this equation.

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Replacing F by E times Q, we get that the work done is equal to E times Q times S.

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Next, let's rearrange this second equation from subtopic 5.1 to give us W equals 2 V times Q.

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Since both EQS and VQ are equal to the work done, we can write that EQS is equal to V times Q.

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Canceling Q and rearranging gives us that E is equal to V over S.

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This equation shows us that when the electric field is uniform,

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electric field strength can be calculated as change in potential divided by distance moved.

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Let's move on to draw the electric potential versus distance graph between the two plates.

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We'll denote the distance with the letter R and the electric potential as VE.

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We'll take both R and VE to be 0 on the negative plate.

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We'll also assume that the distance between the plates is 5 cm.

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Therefore, the equipotentials that you see on the diagram

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have a distance of 1 cm between them.

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Let's plot the graph.

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On the negative plate, where R is 0, VE is also 0.

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At a distance of 1 cm, the electric potential is 1 volt.

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At 2 cm, it's 2 volts, 3 cm, 3 volts, 4 cm, 4 volts and 5 cm, 5 volts.

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The last dot represents the positive plate.

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Connecting the dots, we get a straight line.

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Let's examine the gradient of this line.

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Since we have VE on the vertical and R on the horizontal axis,

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the gradient is equal to change in VE, so delta VE, divided by change in R, so delta R.

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Let's link this gradient to the electric field strength equation.

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Since we have a straight line, we can simply calculate the gradient

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by picking the first and the last point on the graph.

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These two points represent the two plates.

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So in this case, delta VE is equal to the potential difference between the plates,

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so V, and delta R is equal to the distance between the plates, so S.

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Therefore, we can say that delta VE over delta R is equal to the electric field strength.

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There is a final small change that we need to introduce here to be fully correct,

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and this is adding a negative sign to the equation, giving us that E is equal to

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negative delta VE divided by delta R. The reason for the negative sign is that the

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direction of the electric field is opposite to how the potential of a positive charge varies.

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In other words, as the positive charge is moved opposite the electric field,

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it gains potential energy, and as it moves in the direction of the electric field,

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it loses potential energy. Of course, a similar reasoning can also be applied to a negative charge,

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but we won't go into these details here.

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Delta VE over delta R is called a potential gradient, therefore the electric field strength

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is equal to the negative of the potential gradient.

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The formula that shows this connection between electric field strength and the potential gradient

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can be found in the IB Physics data booklet, and here are the variables that are present in this

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formula. Next, let's see how this connection works when the electric field is not uniform.

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A non-uniform electric field that we learned about is one that arises due to a point charge.

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We learned about the electric field due to a point charge in subtopic 5.1.

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I won't repeat the related concepts here again, so feel free to go back to subtopic 5.1

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and revise what we discussed. In this subtopic, we also introduced the following two equations.

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Then, we combine them to get that E is equal to K times Q over R squared.

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This is the equation for the electric field strength

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due to a point charge Q at a distance R from the point charge.

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Earlier in this video, we learned that electric field strength is equal to the negative of the

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potential gradient. The discussion that follows from here will help us derive the equation for

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electric potential at a distance R from a point charge. We'll use calculus in this derivation,

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so don't worry if you find it a bit too complex. You do not need to know this derivation for your

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IB physics exam. So using calculus, we can write the second equation for the electric field strength

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as E is equal to negative dVe over dr. If we equate our two expressions for electric field strength,

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we get a differential equation. We can solve this equation by the separation of variables technique,

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which I won't discuss in detail here, but which gives us that Ve is equal to K times Q over R.

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This is the equation in the IB physics data booklet for electric potential due to a point

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charge, and here are the variables that are present in this formula. Before moving on to

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gravitational fields, let's compare the electric potential versus distance and the electric field

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strength versus distance graphs for a point charge. You can see that the general shape of the

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graphs is similar, but since the potential graph is a 1 over R graph, while the electric field

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strength graph is a 1 over R squared graph, there is a difference in the steepness of the graphs.

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Next, let's move on to gravitational fields. The underlying concepts of different fields in

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physics are very similar. What we learned about electric fields so far in this video

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can also be applied to gravitational fields. Gravitational field strength is equal to the

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negative of the potential gradient, which we can express using this formula. Here are the variables

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that are present in this formula. Delta Vg over delta R is called the potential gradient.

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Combining this formula and the one that we learned in subtopics 6.2 for gravitational field strength

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and using calculus, we can show that Vg is equal to negative g times m over R. Here are the variables

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in this formula. The derivation of these two equations is very similar to what we have seen

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when we derived similar equations for electric fields. Therefore here we won't go into the details

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of where these formulae come from. Just note that gravitational potential Vg is always negative.

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We discussed the reasons for this in subtopic 10.1. Just as we have done for electric fields,

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let's finish up with two graphs. One is gravitational potential versus distance and the other

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is gravitational field strength versus distance for a point mass. Just as for electric fields,

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you can see here that the shape of the graphs is similar and that the main difference is between

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their steepness. Another important feature that we already mentioned before is that as you can see

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gravitational potential is always negative. Let's summarize what we have learned in this video.

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We discussed several concepts, but the main takeaway from here

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are the four formulae that we discovered. Two of these are related to electric fields

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and two are related to gravitational fields. For electric fields, we'll learn the connection between

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field strength and potential gradient and the formula for electric potential due to a point

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charge. For gravitational fields, we connected gravitational field strength with the potential

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gradient and then saw the formula for gravitational potential due to a point mass.

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As a final reminder, let me add that this expression and this expression are called

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potential gradient. This completes our discussion of field strength, potential

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and potential gradient. In the next video, we'll learn about potential energy.