Relationships Between Force, Field, Energy, Potential - Online Tutor, Practice Problems & Exam Prep

Understanding electric forces, fields, potential energy, and potential is crucial in physics. The electric field (E) exerts a force (F) on a charge (q) as described by the equation F^2 = qE. Similarly, potential energy (U) relates to potential (V) through U = qV. The relationship between potential energy and force is given by -ΔU = FΔr, while potential difference (voltage) relates to electric field as -ΔV = EΔr.

Relationships Between Force, Field, Energy, Potential

Alright, guys. So we've seen a lot of stuff so far. We've seen electric forces. We've seen electric fields, potential energies, and potentials. And now we're basically going to put it all together in an awesome formula sheet that's going to be really helpful for your tests. Let's go ahead and check it out. So far, we've seen these four related things, and often it's really hard to keep track of all of them. Which one's which? Which one involves what variables? So this table is going to be really awesome at putting all of these things in perspective together. Let's go ahead and take a look at it.

We've got r2. We've got some equations involving r squared. Some that involve single r's. Some that involve multiple charges, and some that involve single charges. So basically, let's take it from the bottom. We know that this electric field here is a force field. And the best way that we can think about this is that it is a charge that sets up a field that another charge will feel and experience a force from. Now, I just want to point out really quickly that I know I'm using little q's here when I've used big Q's, and I know that's potentially might be confusing. But the thing is that it won't matter. You just have to be sure that on tests force. Because the reality is that some of your professors might use q1q2, force. Because the reality is that some of your professors might use q1, q2. Some of your professors might use big Q's and little q's. It's up to you to decide and figure out which one is the producing charge and which one's the feeling charge.

In any case, we know that a charge that it produces a field that is e, has a resulting force on a feeling charge. The relationship between these two formulas is f=q⁢e. So if one charge produces an electric field, and another charge, which is this little q right here, feels that electric field, it's going to have a force.

Similarly, we know that a charge will also produce something called a potential. And this potential is basically just an energy field. Instead of telling other charges how much force to feel, it's basically telling other charges how much energy to have. And we know that this relationship between these two formulas is given as u=q⁢v.

Now, what we haven't seen yet is we haven't seen the relationships between these two formulas, f and u. So basically, the negative of the potential energy difference is going to be f⁢∇r. Most of the time, you won't see this as delta r. A lot of the times you'll see this as delta x instead. But I want to point out that I'm using this delta r because it sort of helps reinforce the relationship between these r's right here. But ultimately, these are distance variables.

And we have a similar relationship between the electric field and the electric potential, and that's that the negative of the potential difference here is just going to be equal to e⁢∇r, or sometimes, most of the time, you'll see this as delta x. Again, just reinforcing the fact that it's just a distance variable.

The last thing I want to do is point out that these two quantities, these deltas, actually have special names. Remember that this delta v is defined as the voltage. It's the potential difference between these two points. That's delta v right here. That's the voltage. And this delta u over here, the negative in the change of the potential energy seen so far. Work, potential energy, electric force, electric field, all of those things. Alright?

So basically, that's it for this video. Go ahead and print out this page, take it to your exam, and you'll be good to go. Alright?

Potential at Center of Charges in a Square

Alright, guys. We're going to work this one out together. So we have the square of charges right here. They're not all equal charges, but what we're supposed to do is find the potential at the center of this arrangement right here. So, just right off the bat: potential. Which formula is that? Potential is a formula of 1q and 1r. So, in other words, the potential is k times the producing charge at some distance right here. And we're supposed to find out what the total potential at this point right here is. This is our point of interest. So, basically, at the center of this arrangement right here, all we need to do is figure out what the distances are to each one of these charges that are on the corners. So, one of the things we can do is sort of break this up into a little triangle, like a mini triangle. And basically, we've cut this thing in half. So we know that this distance, which is half, is going to be 2.5 millimeters. And this is also going to be 2 and a half millimeters. So, that means we can actually figure out by using the hypotenuse in the Pythagorean theorem, that this is basically 3.54 millimeters. Right? And that's just something we get from the Pythagorean theorem.

So we know that this r distance right here, which is 3.54 millimeters, is going to be 3.54 times 10 to the minus 3 meters. So basically, the idea is that all of these charges are going to be producing potentials at the center. And these charges, these potentials that they produce are going to be scalars, which means that we can just add them up together. So the total potential - that's one thing I could possibly write - is just V_total is going to be V, the potential from the 2 nanocoulomb charge plus the potential from the 1.5 nanocoulomb charge, to the negative 1.5, and then so on and so forth. Right? You can basically just add up all of those potentials together.

So that means that the potentials here, V_total, you would get by using kq/r. And the only thing that's changing is what, just depending on the producing charge that you're putting inside that equation. But if you add all of these things up together, what ends up happening is that this k and this r end up being greatest common factors. So, one of the shortcuts that you can use is say that this is going to be k/r and then basically, just add up all of the charges that contribute a potential, Q₁, Q₂, Q₃, and then so on and so forth. Right? Because, basically, you could just add those things up together. You don't have to keep sticking k/r into your calculator over and over.

So what we can do is that the total potential is going to be 8.99 times 10 to the 9th. Now, we've got the distance between all of these charges and the center is always going to be 3.54 times 10 to the minus 3. And now, we just add the charges up together commutatively. It doesn't matter which order you do it in. So that means I could do 2 nanocoulombs minus 3 nanocoulombs. Let me see. Minus 3 nanocoulombs and then plus 1 nanocoulomb and then we've got minus 1.5 nanocoulombs. Let me scoot down. So, basically, what happens here is if you close off this parenthesis, this 2, the negative 3, and the plus 1 are all going to cancel out to 0. And all you have to do is just do one multiplication here. 8.99 times 10 to the 9th is k divided by r and then times this negative 1.5 nanocoulombs, which is 10 to the negative 9 by the way. So, don't forget to stick this in as 10 to the minus 9. And you should get that the potential is equal to negative 3.81 times 10 to the 3rd, and that's going to be in volts. So, that's negative by the way. And I just want to point out the fact that you could only do the shortcut here because all of these distances, if you know how to cancel these things out and take these shortcuts, then it's going to be a great tool for you to use. Let me know if you guys have any questions in the comments, and I'll see you in the next one.

A -2C charge lies at rest. (a) What is the potential difference between point A, which is 1.5m from the charge, and point B, which is 4m from the charge? (b) What would the work on a 4C charge be to move it from A to B?

• A. (a) −6.9×10⁹ V; (b) +2.7×10¹⁰ J
• B. (a) +6.9×10⁹ V; (b) −2.7×10¹⁰ J
• C. (a) −7.5×10⁹ V; (b) +3.0×10¹⁰ J
• D. (a) +7.5×10⁹ V; (b) −3.0×10¹⁰ J
• E. (a) −1.6×10¹⁰ V; (b) +6.4×10¹⁰ J
• F. (a) +1.6×10¹⁰ V; (b) −6.4×10¹⁰ J

Potential Difference Between Two Charges

All right, guys. In this example, we're upping the ante. Now we're going to figure out what the potential difference is, not just to one charge, but due to 2 charges. So we've got this 5 nano nanocoulomb charge and a negative 3 nano coulomb charge on the same line. So I'm just gonna draw a quick little sketch right here. I've got this 5 nano coulomb charge, negative 3 nano coulomb charge, and I know that they're on the same line and that distance is 6 millimeters. Now, in part a, what I'm asked for is I'm asked for the potential directly between them. So in other words, this is gonna be the potential at VA. So let me do a better job at sort of highlighting that. So that is gonna be the potential at VA. Alright? So let's go ahead and do work that out.

We've got the potential at VA is basically gonna be the potential due to both of these charges at this location. But one of the things we know is that if this is half of the distance between these two charges, then that means that this distance right here, this 3 millimeters, is actually gonna be the same for both of them. So in other words, we've got the potential at point a is gonna be the potential due to the 5 nanocoulomb charge plus the potential due to the negative 3 nanocoulomb charge. And we're just gonna add those things up together because they're scalars. So we have that the potential difference over here is just gonna be k⋅5ra1+k⋅3ra2. But one of the things we notice is that we have the same exact Ks and RAs for both of these things.

So in other words, since both of these are the same, we can actually use a little shortcut that we've used before in reducing these potentials down to, like, even simpler forms. We have the potentials basically just equal to kra⋅(5-3). And it's only because we have symmetry. It's only because these things are the same distances that we can actually use this rule. So in other words, that the potential at point a is gonna be 8.999×109, and then this is gonna be divided by the distance. We have to be careful because this is 3 millimeters, which means it's 0.003, and then we're just gonna add up the charges. We have 5 nanocoulombs minus the 3 nanocoulombs.

So that means that the result here so both of these things together are just gonna add up to 2 nanocoulomb charges. Right? So you're just gonna add in both of these charges, and you get 2 nanocoulombs, which by the way can be represented by 2×10-9. That's gonna be in coulombs. Right? So that means the potential at point a is just going to be Let's see. I got 6×103, that's in volts. Alright? So we can use that shortcut for this case right here.

So in part b now, in part b, we're supposed to figure out what is the potential at this point b, which is now halfway between the charges, but now it's some extra distance above that line. So in other words, point b is somewhere over here, and we're supposed to be figuring out what is the potential at this point. So basically what we need to do is now that we have a different point, we actually have to calculate what the distance is to both of these charges. Now fortunately, what we've got here, is we've got a 3, and we know that this line right here is 4 millimeters above, so we can recognize this as a 3, 4, 5 triangle. If you wanted to figure out this r distance, if you didn't know if it's a 3 4 5, you could always just use the Pythagorean theorem. But in any case, we've got 5 millimeters right here for this distance and this RB is actually gonna be the same for both of these things. Right? Because it's symmetrically placed around around this, this axis.

So you can kinda use the same shortcut that we did here, but now we're just gonna sort of like do it a little bit quicker. So this VB is just gonna be krb⋅(5-3). And if you write it all out if we write it all out, we get that the potential at point b is 8.999×109. And now we've got the distance which is 0.005, and now we've got 2×10-9. And if you work that out, you should get the potential at this point is equal to 3.6×103, and that's in volts. So that is part a and part b. And now the last part over here, which I wanna do, let's say over here, is I need to figure out what the work that's done on a 1 nano coulomb charge. And let's see. We've got from the first point to the second point. In other words, we're going from point a over to point b. And we know that a work that's done on a feeling charge through a potential difference is negative q, which sets the feeling charge times the potential difference at this point, and this is the potential from a to b. So in other words, what happens is the work that's done is equal to negative q times the difference in the final minus initial potentials, vb minus va. If it had set the second point to the first, then we would actually reverse that. So it's very important that you do the right step here.

Alright. So we've got the work that's done. This is gonna be negative. We've got 1×10-9. That's the feeling charge that we have. This negative one_nanocoulomb charge is this feeling charge over here. And then we've got the potential difference. So v b was equal to 3.6×103. V a, was 6×103. And if you work all that out with your calculator By the way, you could have actually just figured out what the potential difference is. Just like I actually gotten the subtraction, and then you would just plug that in here. But I'm just doing it sort of like the expanded way. And if you work this all out, you should get a work that's done, which is equal to 2.4×10-6, and that's in joules. So that's the work done."]

Distance to Stop a Point Charge

Hey, guys. Hopefully, you were able to take a look at this one on your own. Maybe try it out. But this is going to be a little bit different than the problems that we've seen before. We've got a 5 gram, 3 microcoulomb charge. It has some initial speed, and it's moving away from some negative charge. And then, basically, we're supposed to figure out how far can this thing travel before it stops. So I want to figure out just sort of like what's going on here. So let's go ahead and draw a quick diagram. Now, I've got this negative 5 microcoulomb charge, and then I've got another 3 coulomb or 3 microcoulomb charge. I know the mass of this charge is 5 grams, and I know the initial speed of this is equal to 20 meters per second. So what ends up happening is that as this positive charge is moving away from this negative charge, the force on it wants to pull it back towards the negative charge because these two things are opposite charges. Opposites attract. Right? Now I know what the initial distance is between these two objects, so that our initial is equal to Let's see what this What is that? Five centimeters? Yeah. It's 5 centimeters. So what ends up happening is that as this thing is flying out with some velocity, the force is eventually going to bring it to a stop at some later distance over here. Now, I know at this point be while it stops, the final velocity is going to be 0. And basically, what I need to figure out is what is the horizontal distance, or just horizontal, I'm guessing. What is this distance here that it can travel before it actually stops at this point? So if I can kinda think about this, this will be the final distance, \( r_f \). So that means that this \( x \) distance, if I can write it as an equation, is basically just going to be, I've got \( x \) is equal to \( r_{\text{final}} - r_{\text{initial}} \). And so this really is my target variable. How far can it travel before stopping? I know what the initial distance is. Now I just need to figure out what the final distance is. So in all these cases, I've got this final distance, initial distance. I've got speeds, masses, and I'm also talking about energy. So what I need to use is I actually need to use energy conservation in order to solve this problem. So what I'm going to do is look at all of my energies in the before and after case. So we know that our energy conservation formula is the initial kinetic energy plus any initial potential energy, which we know is going to be electric potential energy. Now there's no work done by non-conservative forces. We know that already. And then that's going to equal the final kinetic energy plus the final electrical potential energy. Okay. So we've got that the work done by non-conservative forces is 0. The electric force is conservative. And we know that when this thing finally stops, is there anything we can cancel out? Well, let's see. We've got an initial kinetic energy because we know the object is moving and we also have an electrical potential energy because we have 2 charges and they're separated by some distance, so there's always potential energy. Now, what happens is when this thing stops at this moment that's right out here, we know that the kinetic energy is equal to 0, but there still is some final potential energy. Okay? So basically, we can go ahead and expand out all of these terms. I know that this initial kinetic energy is going to be \( \frac{1}{2} m v_0^2 \). Now what happens is I've got to do the, the initial, kinetic or sorry. The initial electrical potential energy, which is \( \frac{k q_1 q_2}{r_{\text{initial}}} \), and then that's going to equal the final electric potential energy, \( \frac{kq_1q_2 }{r_{\text{final}}} \). So in other words, I'm actually looking for this \( r_{\text{final}} \) because then I can take this \( r_{\text{final}} \) and plug it back into this formula and \( x \) distance is that it travels before stopping. So that's basically the game. I'm going to just try to figure out what this \( r_{\text{final}} \) is. But what happens is, I can't go around and like start manipulating this by flipping these fractions or flipping these formulas because this is like, this is an addition, and I've got a formula that's a fraction in here. It's going to get really really ugly. Fortunately, what I can do though, is I know what the mass is, I know what the initial velocity is, I know what all of these constants are and including the distance. So rather than try to algebraically manipulate it, just start plugging in numbers for this stuff. Just reduce it down to like a simple number and then worry about the algebra later. So basically, I'm just going to plug in this really long mess right here. We've got 5 grams, so that's going to be 0.005. Now, we've got the initial velocity is 20, so then we're going to square that. Now, you've got plus, and we've got 8.99 times \( 10^9 \). Now the 2 charges that are involved. We've got 3 microcoulombs, so that's going to be 3 times \( 10^{-6} \). And a, let's see. Negative 5 times \( 10^{-6} \). And now you've got the distance, which is point 0, that's point 5. That's in centimeters. Right? So, that's the point 5 meters, and then we've got these formulas right here. We we we know what these charges are. Okay. So, basically, if you plug all of this stuff in, that should equal what \( \frac{k q_1 q_2}{r_{\text{final}}} \) is. And, basically, if you plug all of this stuff in really into your calculator, you're just going to get one big number, or doesn't have to necessarily be big. But this is actually going to be negative 1.7 joules. And that's going to equal \( \frac{k q_1 q_2}{r_{\text{final}}} \). So now, what we can do is now we can actually manipulate what this \( r_{\text{final}} \) is. And basically, what's just going to happen is this \( r_{\text{final}} \) goes up to the top, this negative 1.7 comes down to the bottom, and then we can just go ahead and plug everything in again. So that's why it's going to be easier to do that than actually work through all the algebra in that big step over there. Okay? So we've got the final distance is going to be \( 8.99 \times 10^9 \). Now I've got the 2 charges, which are, let's see. I've got 3 times \( 10^{-6} \), negative 5 times \( 10^{-6} \). By the way, there are some shortcuts that you could take in manipulating some of this algebra, but, yeah. So we've got this right here, and then we've got this negative 1.7. Okay? And now, if you plug all of this stuff in, you should actually get 0.0793. But remember, this is going to be in meters. So what this actually represents is this actually represents 7.93 centimeters. So this either one of these expressions, if you got these if you got this, would be correct, but basically, just this is just a different way of expressing this. But what we can do is we can stick this number. I'm just going to go ahead and stick with the centimeters because we're dealing with centimeters, and then go back and plug that into this formula over here. So now what my \( x \) distance is, what my real target variable, how far can this thing travel before it stops, is going to be 7.93 centimeters minus the initial distance of 5 centimeters. So that just means that this thing travels 2.93 centimeters before stopping. So this right here is actually our final answer, that 2.93 meters/cm. Okay? Let me know if you guys have any questions. This energy conservation stuff can actually become really useful in solving some problems. Sometimes you'll have to do it. So make sure you'\re comfortable with this. Go ahead and watch the video again if you didn't understand everything. Drop me a link, or comments, or a question, anything like that, and I'll see you guys later.

Here’s what students ask on this topic:

What is the relationship between electric force and electric field?

The relationship between electric force (F) and electric field (E) is given by the equation:

$F=qE$

Here, F is the force experienced by a charge q in an electric field E. This means that the force on a charge is directly proportional to the magnitude of the charge and the strength of the electric field. If you know the electric field and the charge, you can easily calculate the force acting on the charge.

How is potential energy related to electric potential?

Potential energy (U) is related to electric potential (V) through the equation:

$U=qV$

In this equation, U is the potential energy, q is the charge, and V is the electric potential. This relationship indicates that the potential energy of a charge in an electric field is the product of the charge and the electric potential at that point. It helps in understanding how much energy a charge has due to its position in the electric field.

What is the formula for the relationship between potential energy difference and force?

The relationship between potential energy difference (ΔU) and force (F) is given by:

$-\Delta U=F\Delta r$

Here, ΔU is the change in potential energy, F is the force, and Δr is the displacement. This equation shows that the negative change in potential energy is equal to the work done by the force over a distance. It is crucial for understanding how energy is transferred in an electric field.

How does potential difference relate to the electric field?

The relationship between potential difference (ΔV) and the electric field (E) is given by:

$-\Delta V=E\Delta r$

In this equation, ΔV is the potential difference, E is the electric field, and Δr is the displacement. This relationship indicates that the negative of the potential difference is equal to the electric field multiplied by the distance over which the field is applied. It helps in understanding how the electric field affects the potential difference between two points.

What is the significance of the voltage in electric fields?

Voltage, or electric potential difference (ΔV), is significant in electric fields because it represents the work needed to move a charge between two points in the field. It is defined as:

$\Delta V=V$_final - V_initial

Voltage is crucial for understanding how much energy per unit charge is available to do work. It is a key concept in circuits and electrostatics, influencing how charges move and how much energy is stored in electric fields.