kum's law going to be the topic of this lesson in my new General Physics playlist which when complete will cover a full year of University algebra based physics now kum's law provides for us the formula for the force between two point charges and we call this the electrostatic force it's one of the four fundamental forces and this is the first place we're encountering it we're going to learn how to treat it as just one other Force to add into our growing repertoire of forces that we can apply to mechanics problems both in one dimension and two Dimensions my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on Chads prep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat so let's Dive Right into koloms law here so F equal K q1 Q2 over r^2 so K is the kulum constant it has a value of 8.9 99 * 10 9 new s perum s so notice we have the absolute value of q1 and the absolute value of Q2 those are the magnitudes of the charges uh that are either going to be attracted to each other or repelling each other and then all over the distance of Separation squared so we kind of got this diagram here so I've got two charges q1 and Q2 they could both be positive they could both be negative or one could be positive and one could be negative now we know that like charges repel so if they're both positive or both negative that's going to be repulsive Force away from each other but if one's positive one's negative we know that unlike charges or opposite charges attract so it would be an attractive Force pulling towards each other all right so uh one thing to note not everybody puts the absolute value brackets on this and they teach that if you plug in q1 and Q2 and they're both positive or both negative that overall the force would come out to be a positive number and in such cases they tell you that that should you should know that that's a repulsive force uh whereas if one of these was positive and one of these was negative then it would come out to be an overall negative number and you should know that when you plug in uh your q1 and Q2 without absolute values and you get an overall negative number for the electrostatic force that is a repulsive Force now I'm not going to teach it that way I'm just going to assume that you know that like charges repel and unlike or opposite charges attract so we'll leave it at that now one thing you should note here so this is very similar to what we saw with the universal formula for gravity so I'm going to compare it to gravity a couple different times in this chapter and the next so if you notice we had FAL G M1 M2 r^2 now with gravity this was the attractive force between objects that have mass whereas for our electrostatic force here with kum's law this is the either attractive or repulsive Force for between objects that have charge instead now notice there's only one kind of mass and so we only ever dealt with uh gravity being an attractive force it was never repulsive but because there's two different types of charges in the universe it turns out one we call positive one we call negative it turns out there's two different uh possible forces one attractive one repulsive that's going to be one key difference so I'll point that out a little bit later so but uh you're used to seeing gravity uh quite a bit but seeing you know electrons uh you know following a force and things of the sort may not be the most common thing to you so when applicable I will relate often times this electrostatic force back to gravity in a couple key places noting one key difference with the possibility of either attractive or or repulsive forces all right so a couple key things to see here so if you look we can see that this electrostatic force is proportional to the magnitude of q1 that electrostatic force is proportional to the magnitude of Q2 if I double the the magnitude of q1 the overall electrostatic force is going to double if I double the magnitude of Q2 the overall force between the two again would double and if I double the magnitude of both charges then the overall force between them is going to quadruple now similar fashion so that force is inversely proportional to the distance of Separation squared so if I double the distance between these two charges so the overall force is not going to go down by a factor of two but by a factor of 2^ squar by a factor of four instead and again that's similar to what we saw with gravity another one of our inverse Square laws and from here we're ready to dive in to do some plugin and chug in so in this diagram we've got up here is as good as any so let's take a look at the first question on here and try to make the numbers nice where possible but we will need your calculator a little bit in this lesson so first question here says two point charges each of magnitude positive 1.0 Ks are 0.1 M apart what is the magnitude of the force experienced by either one of the charges so here we've got positive 1.0 kums and here positive 1.0 kums and here we've got a distance of separation of 0.10 M all right so so not so bad to calculate here so just do some pluggin and chugging FAL K again * the magnitude of q1 * the magnitude of Q2 all over R 2 K is 8.99 * 10 9th and again that's Newton m squar per Kum squared you can see how the math is going to work out here or at least how the units are going to work out so once we multiply q1 and Q2 the ks will end up cancel at least a Kum squared in this case will cancel uh and these were both one Kum so times 1.0 kstimes 1.0 KS and then all over the distance of Separation squared in SI units of meter so 01 M and don't forget to square it and this we can do in our head and I chose some nice numbers here we're going to find out that you know one Kum is actually a fairly significant amount of charge in fact it's a ginormous amount of charge so but it makes the math nice for this first example uh in this case so 8.99 * 10 9th * 1 * 1 is still 8.99 * 10 9th and dividing by 0.1 is the same thing as multiplying by 10 so dividing by .1 squ is the same thing as multiplying by 10^ squ or multiplying by 100 so we can do this one in our head and in this case it's just going to be 8.99 * 10 7th and if you look here the meter squar cancel the kums squared cancel and you're just left with an answer in Newton as it should be so 8.99 * 10 the 7th Newtons now again one Kum is a phenomenal amount of charge and you're not likely to actually encounter a problem with actual point charges having a value of 1.0 K that's actually pretty ridiculous so we're going to redo this problem revising the numbers just a little bit to resemble something you're a little more likely to see all right so swapping these out a little bit so two point charges each of magnitude 1.0 micrum so we'll switch this out to micro kums are 0.1 M apart so we'll leave that alone what is the magnitude of the force experienced by either of the point charges and so in this case where this is going to change this just a little bit instead of one Kum here so we can't plug in micrum we got to put this still in kums but one micrum is 1 * 10 -6 K so I'll change our calculation now again technically you could probably still do this in your head so 8.99 * 10 9th * 1 * 10 6 be 8.99 * 10 3r * an additional 1 * 106 would be 8.99 * 10 -3rd and then again dividing by 0.1 squ the same thing as multiplying by 10^ squar so multiplying by 100 which is going to get you 8.99 * 10 to the1 newtons now or just simply 0.899 uh Newtons cool so I just wanted you to see this both kind of in a a simpler fashion where the math was going to be just a teeny bit easier so so the truth is you can still do this in your head but these are now with micrum or nanocs or pums those are going to be magnitudes of charges you're much more likely to encounter in a problem you're likely to see so the next application of kum's law here is going to end up being a two-dimensional problem and we've got two different point charges acting on a third Point charge and to find the resultant Force we've got to find the resultant Vector we'll be adding those together as vectors all right so the question says what is the magnitude and direction of the net electrostatic force experienced by the positive 2.0 micrum charge in the following diagram and the diagram is definitely included on the study gu is part of the question here so this is part of the problem uh in this case we can draw a really quick free body diagram here for the positive 2.0 micrum charge and uh it's experiencing a force due to this negative 1 micrum charge and because uh the charge of interest is positive and this other one is negative they are unlike or opposite charges and there's attractive force between them and so for the two micrum charge that's going to point to the right now if we cared for the negative 1 micrum charge that would be a force directed to the left so when you do kum's law you're actually calculating the magnitude of the force experienced by either of the point charges so but they will be equal in magnitude opposite in direction to satisfy Newton's third law and so in this case again for the positive charge is direct to the right but for this negative charge it would be directed to the left if we cared all right now our positive two micrum point charge is also feeling a force due to this positive 4.0 microcolon Point charge and they are like charges so it's going to be a repulsive force and so for the 2 micrum charge that's going to be directed up and again if we carried for the positive 4.0 micrum charge that would be directed down again equal in magnitude opposite in direction and so we can see there's really two forces acting on our positive 2.0 micrum charge and the resultant Vector in adding them in to tail fashion is going to be somewhere in between so up here in the first quadrant if you will all right so but we definitely have to add these as vectors so we're going to calculate the force using kums law for both and then add up the resultant Vector all right so let's start with the X Direction here so in this case your force is going to be kqq over R 2 so 8.99 * 10 9th Newton m s per K squar time q and we got 2 * 10- 6 Kum so and 1 * 10- 6 K and notice because I know we're plugging in absolute values I didn't bother to carry the negative sign through it's going to go away anyways and we already figured out that it's directed to the right based on the fact that opposite charges attract all right then all over the distance of Separation squared and one micron is 1 * 10 -6 M and definitely don't forget to square it one of the more common mistakes to make now this is probably one you could do in your head but uh you can definitely let your calculator do the work for you as well but you might see that this 10- 6^ s could cancel out both of these 10us 6es and you're really just left with 8.99 * 10 9 * 2 and really times 1 but anything times 1 is itself so that's ultimately what this is going to come down to so 8.99 * 10 9th * 2 e -6 * 1 E -6 and divided 1 E -6 sared don't forget to square that and it comes out to just as if we'd multiplied 8.99 * * 2 comes out to 1798 * 10 10th I'll write that over here actually all right that is a rather large number all right let's move on here and now find the force between the positive2 micrum and positive4 micrum charge uh set up kum's law once again 8.99 * 10 9th Newton m s per k s * 2 * 10 -6 K * 4 * 10-6 Kum and once again all over one micron micrometer micrometer however you want to say it 1 * 10 -6 M and once again do not forget to square that number and similar to the last one you're ultimately going to end up with 8.99 x 10 9th * 2 * 4 because the 10 the6 parts will cancel but by all means let's your calculator do the math and for me I'm actually just going to put in the last answer I just stuck in and just change some numbers here I'm going to change the two * 10 no actually I'm not going to change that one leave that one alone but I'll change the 1 * 106 we had in before and make that four * 106 so you can if you don't know how to use your TI calculator if you're using one you can use the second function enter to put in the last entry which is just what I did there so and from here oop I put the four in the wrong place let's get that in the right place so four goes there denominator is still 1 * 106 s and we're going to get 7192 times 10 to the something Newtons and I got to count some zeros here so one two three four five six 7 8 9 10 still times 10 to the 10th cuz 9 * 8 still would have been y * 10 10th 72 * 10 10th or 10 9th which is 7192 * 10 10th Newtons in this case and again just as a reminder this is in the positive X Direction This is in the positive y direction uh which makes it convenient so normally if these weren't purely in the x or purely in the Y we'd have to break them up into components using SS and cosiness so in this case this is all X this is all Y and we can go straight to our Pythagorean theorem to get the magnitude of the resultant vector and so in this case we're we're going to have 1 798 * 10 10th squared plus 7192 * 10 10th squared we'll take the square root to get the result the magnitude of the resultant Vector here in this case and so again 1.7 98 * 10 10th squar + 7192 * 10 10th squared equals and then square root of that answer we're going to get 7.41 uh in fact I want to round this to two Sig fig so just 7.4 * 10 the 1 2 3 4 5 6 7 8 9 10 still so 7.4 * 10 the 10th Newtons now that's our our magnitude of our Force now we still need a direction you might recall that tangent Theta is going to equal the Y component over the X component in this case that y component was again the 7 1 192 * 10 10th over the 1.7 98 * 10 10th and we'll take the inverse tangent to get the angle all right so 7192 * 10 10th and if you realize it you could just drop the 10 the 10th part but I'll just keep it in so divided by 1798 * 10 10th equals then I'll take the inverse tangent of that last answer so by the way this came out to exactly four and then taking the inverse tangent is going to get us 75.9 6 degrees which I'll just round to 76° so cool and that makes sense so we we figured it should be bigger than 45° since the force uh due to the four micrum charge is going to be bigger than the force due to the Nega 1 micrum charge and so it should be pointing more in the y direction Less in the X Direction definitely should have been an angle bigger than 45° obviously still less than 90 so but there's your answer so force is 7.4 time 10 the 10th Newtons at an angle of 76 degrees so this last question is also going to be a two-dimensional question but it's going to be a two-dimensional mechanics question so we're going to draw a free body diagram we're going to do some of the forces or net force in the X Direction net force in the y direction and then solve that system of equations so the question says two small metallic spheres each of mass 1.0 G are suspended by very light strings I strings that are effectively massless from a common Point as shown in the following figure the Spheres are given the same electric charge and it is found that they come to equilibrium when each string is at an angle of 30.0 De with the vertical if each string is 10.0 cm long what is the magnitude of the charge on each sphere so because these two spheres have exactly the same charge either both positive both negative it's going to be a repulsive force and so they're pushing each other apart and it found turns out they come to equilibrium when the angle with the vertical on either side here equals 30.0 de so we're going to set up free body diagram set up again some of the equations or some of the forces net force equations and then solve the system till we can ultimately solve for Q here now when we set up kum's law we usually have q1 Q2 but because these have exactly the same magnitude of charge I'm just going to generically use Q which makes this just a little bit easier problem so all right let's start with the free body diagram here and we can choose for either one of these since they're identical and I'll do the one on the left here and so in this case we have the weight pointing down and that's mg we have a tension pointing up the string here called t and then a repulsive Force pointing to the left away from the other charge which should be K q^2 Over r s the kulic force all right so we can definitely see this is a two-dimensional problem so we've got components of different forces in the X and Y directions so we have to break this up a little bit and so we'll have the sum of the forces in the X Direction and in this case in the X Direction we've got a component of tension and then the culic force the T the component of is going to point in the positive uh X Direction and in this case that component is going to be T and in this case if we kind of make finish the right triangle here so if this angle up top is 30° then this angle down here is going to be 60° right here and that's typically the angle with respect to the x-axis that we used and so in this case the component of the tension that points in the positive X direction is T cosine the adjacent side here 60 so T cosine 60° and pointing the opposite direction in the negative X direction is the kulic so minus k q ^2 over R 2 and again we're in equilibrium so that adds up to zero and then the y direction some of the forces this case once again in in the positive y direction we've got a component of tension that's going to be t s of 60° so and then pointing downward is the weight so minus m G and once again we're in equilibrium so that adds up to zero there's our two equations and ultimately we're going to want to solve for Q now a couple things we don't have the distance of Separation just yet but we can figure it out so one we know the length of these uh strings is 10.0 cm and so we look at our right triangle this is adjacent to the 60° angle that's the hypotenuse for 10 cm so this is going to equal 10 cm * cine of 60° or S of 30° if you want to look at the opposite the 30° side either way whether we use cosine of 60 or S of 30 it's equal to 12 and 10 * 1 12 is 5 cm and so you've got 5 cm here but you're going to have an additional 5 cm right here and so the distance of separ separation is just 10 cm which makes the math a little bit easier here and setting up the equation is just a little bit easier as well so we do know the distance of Separation here so but to solve for Q it would mean we also have to know uh the tension which we don't yet know so but we'll use the other one to solve for tension here so in this case we'll have tension Time s of 60° is going to equal mg so our tension is going to equal mg over the sign of 60° now we can do some plugging and chugging here so tension is going to equal our Mass which again was one gram but this has got to be in kilogram so that's 1 * 10 -3 kg * 9.8 m/s squared all over the S of 60° let's get that tension let our calculator do the heavy lifting for us all right so 1 * 103 * 9.8 / the S of 60 gets us 0.0113 Newtons all right so there's our tension and we can take that tension and plug it right back in here so to solve for Q and so I'll rearrange this just a little bit for just a second uh we've got T cosine of 60° it's going to equal I'll add this to the other side k q ^2 over r^ 2 T cosine 60° time R 2 over K is going to equal Q ^2 and then we'll take the square root of both both sides and be done all right so that tension was again 0.0113 Newtons time cosine of 60 which happens again to equal 12 times the distance of Separation squared in 10 cm is 0.1 M don't forget to square it and then all over K 8.99 * 10 9th newon me squ per Kum squared cool and then we'll take the square root to get Q here cool we could have done this in steps but you know sometimes it's convenient just to kind of solve for your variable before you plug all the numbers in uh like so so in this case 0.0113 and it's the last answer in my calculator I kind of leave that there times the cosine of 60 * .1 squared and divid 8.99 * 10 9th that's going to get me 6.29 * 10- 15 we'll take the square root of that and get 7.93 * 10us 8 uh and in the end we're going to want looks like two sigfigs so I'll round that to 7.9 * 10us 8 Kum cool you could also look at this as being equal to 79 Nano K as well or 79 * 10- 9 K if you're taking a multiple choice test you might see get represented either way the only point I bring that up it's not uncommon to see charges again uh expressed either in micrum or nanocs or pums whichever happens to be kind of nearby and most convenient if you found this lesson helpful 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