# Coulomb's Law: Example

by Jared Rovny

Comments
My Notes
• Required.
Learning Material 2
• PDF
Slides Electrostatics1 Physics.pdf
• PDF
Download Lecture Overview
Report mistake
Transcript

00:01 We can see an example of exactly how we would add these forces together.

00:04 How we would use superposition at an example like this one.

00:07 Suppose we have an electron, it's equidistant from 2 protons, meaning the distance between them is the same and there is one micron away which is 1 micrometer from their midpoint.

00:18 The question is if the 2 protons are themselves 2 microns apart from each other, what is the magnitude and the direction of the force acting on the electron? You can go ahead and try to use Coulomb's law as we've already introduced it while also keeping in mind that these are vector quantities that they have direction to them, to try to solve this problem and it would be very instructive to give it a shot and see what you arrive at and see if you can get the same answer that we're going to get here.

00:44 Hopefully what you did looked a little something like this. We have these 2 protons.

00:50 It's very important to get the geometry of this one right which can be tricky and we have a distance between them which we said was 2 microns or 2 micrometers.

01:00 We also have an electron over here which again is much smaller which we said is equidistant from the 2 protons, so we have a distance between these 2, which is the same, which we also said had a particular distance away from their midpoint.

01:14 So, this distance here is 1 micrometer while again, this vertical distance between the 2 protons is 2 micrometers.

01:22 So, this electron it's important to know also that we sometimes symbolize it as this, e with a minus sign, this is just a way to represent that we're talking about an electron and then we have these 2 positive forces from each proton.

01:35 So, let's go ahead and think about the vectors that are at play here as well as the direction.

01:41 So, we have a force on this electron acting towards the upper proton.

01:46 So, we have a force to the upper right where we know the dimensions of this triangle, we know that this is a distance of 1 micron and we also know that this is a distance of 1 micron.

01:57 It's 1 because we know this entire distance is 2 and we're only talking about half that distance.

02:02 Similarly, our electron is also experiencing a force to the bottom right and again we know the dimensions of this triangle as being exactly the same.

02:12 So, the question is with these 2 vectors can we write down what the forces are? What we'll do is say that this is maybe the force from the 1st one and maybe this is the force from the 2nd proton.

02:23 So, we could label these, maybe call this one 1 and call this one 2 and then we can write down what these forces are.

02:29 The sum of the 2 forces where again we have to make sure that we treat these 2 forces also as vectors.

02:36 So, if we wanna be sure to add this properly what we should do is go ahead and find out what these components are of the vectors and write out the vectors very clearly.

02:43 So, for each of these suppose we had some angle here, theta.

02:47 So, this is just a way to do this by direct analogy with things that we've done in the past since we have a triangle here.

02:54 The reason we could do this and write it this way is because we already know, and let's get rid of these for just a moment, that the horizontal component of a force if we have an angle theta here will be the force times the cosine of that angle while the vertical component will be the force times the sine of that angle, and we have the same thing for this lower one.

03:14 We have force 2 times the cosine of theta and then we have force 1 times the sine of theta but now we have to be careful about our directions.

03:23 In both of these cases, the forces that are horizontal are acting to the right whereas one of the protons is trying to pull the electron up and the other one is trying to pull it down.

03:32 So, we'll have to be careful with those signs as well.

03:36 So, let's write down what these vectors are.

03:37 So, the first force is a vector. We can write the x and y components so let's introduce a coordinate system, I'll put it over here.

03:44 We have the positive x direction maybe and the positive y direction.

03:49 So, in this case, the force from the first proton would be in the horizontal, the x direction, F1 times the cosine of theta and in the vertical direction, F1 times the sine of theta and both of these are positive by our convention here, by our coordinate system. Whereas F2 is slightly different.

04:07 It has the same x component, F times the cosine of theta but it has the y component.

04:14 This guy is pointing downwards to the negative direction.

04:17 So, we have minus F2 times the sine of theta.

04:20 So we can see that the total force which will be the sum of these 2 forces will be F times the cosine of theta, times, plus rather, F times the cosine of theta.

04:34 So, the x component is 2 times the force times the cosine of theta.

04:38 The y component will be the exact opposite.

04:41 One of them is positive while one is negative so we need to compare these magnitudes now because you can see I've already assumed these forces are the same and the reason I did that is because whatever distance this is, is the same distance as this and also the charges are the same, this one's negative and both of these are the same positive value.

04:58 So, the force, this force F1 and this force F2 are the same so let's just note that here, that F1 has to be equal to F2.

05:07 We'll be using that as we go. So, the total force in the x direction is twice this force, F1 or F2, times the cosine of theta where the difference between these two will be 0 since they're the same but one has a minus sign.

05:21 So, this is our total force as a vector that this electron is experiencing.

05:27 One way to write this is also to, to simply write the x component of this force and say the x component is twice the total force times the cosine of the angle between these.

05:39 The last thing we need to do now is find what this F is. What is the force? So, now we're ready to use Coulomb's law.

05:45 We know that the force is equal to k times the product of the two charges q1 and q2, divided by the distance between those two charges squared and so by writing this force what we have, I'll write it over here so we have a little more room.

06:04 We have 2 times k times the product of the two charges divided by the distance between them squared.

06:12 We finally have the cosine of theta and what I'm gonna do here is now that we have the force written out but we already know the direction.

06:22 We already know the direction is in the positive direction to the right.

06:26 This will be the direction of the net force on these objects.

06:29 Since we already know what the direction is going to be let's just calculate the magnitude of the force in the x direction and not worry about where the signs coming from.

06:37 The reason I'm doing this is we don't wanna confuse ourselves by putting both electric charges in here and then see maybe minus signs popping up and not being sure what direction the force is acting because we already know what direction the force is acting.

06:49 So, let's just say we know the direction.

06:53 So, now all we have to do is calculate what this quantity is.

06:58 Let's do a little bit of simplifying and say that this is equal to 2 times k.

07:06 These charges, q1 and q2, we know that q1 is the charge of an electron which is minus e, we know the charge q2 is the charge of a proton which is plus e and therefore multiplying these two q's together we have e squared.

07:23 The reason I'm not putting the minus sign here is again we're taking the magnitude of this expression because we already know the direction that it should be in.

07:30 Now all we have to do is divide by the distance between these two squared, so this is our r in this equation, in this scenario rather.

07:38 We know what r is because we have a triangle, very simply one more time, we see it right here.

07:43 If this is the electron and we have a proton up here, we have a triangle with 1 micrometer and 1 micrometer in each of these cases.

07:50 So, this diagonal we could find very quickly by the Pythagorean Theorem using c squared is a squared plus b squared to be the square root of 2 micrometers.

08:00 We have to be careful with our units here because we're going to write everything in SI units.

08:05 So we have the square root of 2 micrometers is the same as 1 times 10 to the minus 6th meters.

08:11 So this will be the distance between these two, square root of 2. Sorry, this is the square root of 2, times 10 the minus 6th. Let's be careful with our values here, square root of 2.

08:23 Times 10 to the minus 6th. And this quantity is squared.

08:31 Lastly, we also have this cosine of theta and this is our entire expression.

08:37 So, let's plug in some numbers. We have that the...

08:42 I'll catalog these numbers over here just so we don't get in our own way.

08:45 We already knew that this e is 1.6 times 10 to the minus 19th coulombs.

08:52 We know what this electric constant k zero is.

08:57 This is 9 times 10 to the 9th Newton meter squared per coulombs squared and we also know the distance which we just found to be up here.

09:10 R is the square root of 2 times 10 to the minus 6th meters.

09:15 So we'll keep our values sort of separated over here.

09:18 There's one last thing we need which is the cosine of theta.

09:21 So, what is that' This is our triangle here with this angle theta down here.

09:26 So, the cosine of this angle we can think of it in one or two ways.

09:29 One, is to recognize that the legs of this right triangle are both the same size.

09:33 So, this angle has to be a 45 degree angle and we know the cosine of a 45 degree angle is 1 over the square root of 2 or the square root of 2 over 2.

09:42 The other way we could do this is just by knowing that the cosine of an angle in a right triangle is equal to the length of the adjacent side divided by the length of the hypotenuse.

09:52 In that case we would also get 1 divided by the square root of 2 which is mathematically equivalent to the square root of 2 over 2.

10:00 So, whichever way you do it, here's the value for the cosine of theta and now we're ready to plug in all of these numbers.

10:07 So, we have that the magnitude of the force just in the x direction since we know the y direction is zero, is equal to, doing a little bit of simplifying here, 9 times 10 to the 9th, let's keep all of our units just for a moment just so we're sure we get this correct, times the charge 1.6 times 10 to the minus 19th coulombs squared and then left over if we were careful with all of our cancellations.

10:36 We have square root of 2 over the distance squared, square root of 2 times 10 to the minus 6 squared.

10:44 You might notice one thing that's different is that this had a factor of 2 up front.

10:49 The reason that's gone is from our cosine term.

10:52 Instead of writing square root of 2 over 2 which is equal to cosine, I've cancelled this denominator's 2 with the numerator's 2.

10:59 So this should be the total expression that you get.

11:02 What we're going to do to make this very simple is first of all, to make sure our units work.

11:07 We have coulombs squared down here and coulombs squared up here, so that's great.

11:10 Our distance down here is still in meters.

11:13 So we have meter squared down here and meter squared up here, so that's good as well.

11:17 This will leave us with units of Newtons which is a unit of force so that makes sure that all of this works out well.

11:22 What we're going to do one more time to make this very simple is to gather all of our numbers without their exponents to the left, 9 times 1.6 squared times the number square root of 2 from up here and square root of 2 squared which is 2 from down here and then we're going to put all of our exponents to the right.

11:51 So, we have a 10 to the 9th, we have a 10 to the minus 19th times 2 which is 38, we have a 10 to the 12th, the reason this is positive 12 is that when we multiply this 6 by 2 we get a minus 12 from this negative sign and then we move it to the numerator which makes it a positive 12.

12:14 So, now we have these two terms.

12:16 We have our numbers, our constants, and we also have our actual exponents here.

12:20 Putting these together, hopefully you get an answer something like mine which is approximately equal to 16.3.

12:28 Our exponents here times 10 to the 9th, minus 38 plus 12 is equal to minus 17 and this is in Newtons.

12:41 So, this a very small value for the force experienced by one small charge being pulled towards another small charge.

12:51 We have two charges here and so we had a factor of 2 in the front which we used when we multiplied and we saw that the vertical components of these forces cancelled while the horizontal components of these two forces added to give us twice the force that it would normally experience from one of these protons in the horizontal direction and this is a very common thing so be on the lookout for this often with these vector magnitudes as we saw, some forces will cancel and other ones will add up constructively.

### About the Lecture

The lecture Coulomb's Law: Example by Jared Rovny is from the course Electrostatics.

### Included Quiz Questions

1. 12,000 N
2. 14,000 N
3. 7,000 N
4. 8,000 N
5. 5,000 N
1. 878.9 N
2. 877.6 N
3. 859.3 N
4. 798 N
5. 655.7 N
1. Fg = 3.6 x 10-47 N, Fe = 8.2 x 10-8 N
2. Fg = 3.6 x 10-28 N, Fe = 8.2 x 10-24 N
3. Fg = 8.2 x 10-8 N, Fe = 3.6 x 10-47 N
4. Fg = 2.6 x 10-47 N, Fe = 6.2 x 10-8 N
5. Fg = 6.2 x 10-28 N., Fe = 2.6 x 10-24 N
1. 2.24 x 10^-5
2. 4.24 x 10^-5
3. 2.44 x 10^-5
4. 2.84 x 10^-5
5. 2.94 x 10^-5

### Author of lecture Coulomb's Law: Example ### Customer reviews

(1)
5,0 of 5 stars
 5 Stars 1 4 Stars 0 3 Stars 0 2 Stars 0 1  Star 0

Excellent.
By Karan D. on 14. November 2017 for Coulomb's Law: Example

Jared has concise and clear approach to the subject of physics. A good teacher is very important to understand concepts of physics.