electric Fields 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 the electrostatic force we were introduced to in the last lesson is a field Force just like gravity and it can operate at a distance it is not dependent upon two objects being in contact whatsoever now in addition to dealing with some of the math and some calculations involving electric Fields we're also going to learn how to properly draw electric field lines in this lesson as well 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 In by taking a look at the mathematical relationship between force and electric field so it's a rather simple relationship f equals QE so and the force here is also electrostatic force and we saw that kum's law gives us one expression for the electrostatic force here's another one that just relates it very simply to the electric field at a given point in space uh what do you need in order to fill gravity you need Mass well what do you need to fill electrostatic force you need charge and that becomes part of the equation here as well so the electrostatic force is proportional to the magnitud of the charge that's experiencing the force but also proportional to the magnitude of the electric field at that point in space and if we rearrange this and solve for an expression for the electric field we get force over charge and that's going to let us see that the SI unit for electric field is Newton per Kum all right so you might do some simple plugin and chugging with this if you know the magnitude of electric field at a point in space and you want to know what is the electric uh electric or electrostatic force experienced by some you know charge of some magnitude you simply just multiply the two together to get the magnitude of the electrostatic force that can be rather easy now even if you don't know why there's an electric field at that particular location as long as you know it's magnitude you're good now what causes an electric field well charges do and it might be just one charge it might be a multitude of charges if it's just one charge life is easy we're going to derive a simple expression for what the electric field caused by a point charges but if it's a multitude of charges you might have multiple cont you know contributions to this electric field and you would add them together as vectors so the electric field is a vector you would add them together as vectors so you might have to break them up into X Y components all that but once you knew the overall magnitude and direction of the electric field you could then very quickly calculate the electrostatic force experienced by a charge of given magnitude at that location let's go take a look back at kum's law for just a second we'll have a positive charge and a negative charge and kum's law said that f equals K time the absolute value of Q q1 * the absolute value of Q2 all over the distance of Separation squared and I'm going to write that just a little bit different but that kulic Force that's the electrostatic force experienced by either of the charges and the the force that either one of these FS is equal in magnitude but opposite direction now we know that unlike or opposite charge is attract so for the positive charge it's filling an attractive Force to the right for the negative charge an attractive Force to the left but the magnitude is exactly the same and in either case the magnitude is given by kum's law I'm write that just a little bit different for these I'm going to label so the positive charge is q1 and the negative charge is Q2 and so I'm going to write this just a little bit different so kum's law could also be written as q1 * K * Q2 over r s and over here I'm going to write it as Q2 * K * q1 over R 2 and we want to match this up with our new expression relating electrostatic force and electric field so F = Q E F = Q E when we write it this way the Q so is the Q of the charge that's experiencing the electrostatic force so in this case that would be Q one and over here the force is being felt by Q2 but again point charges not only can feel electrostatic forces they also can cause electrostatic fields that can result in another Point charge feeling an electr force and so here why does q1 feel an electrostatic force well it's because Q2 is causing an electrostatic field to exist at q1's location why does Q2 feel an electrostatic force well it's because q1 is causing electric field to exist at q2's location this gets a little bit confusing that's exactly how it works and in either case though you can see that we can come up with a another simple expression for the electric field caused by a point charge and it's equal to K * Q over R 2 and so it turns out we get this electric field everywhere around a point charge but the further you get away the smaller the magnitude of the electric field is going to become another inverse Square law as we see so if you double the distance away the electric field is going to go down by a factor of four in that case all right so students sometimes get confused about you know who's q1 here and who's Q2 and and how do I you know turn this into QE and things of this sort well again the Q and f equals QE that Q is the the the magnitude of the charge that's feeling the force so and why is it feeling a force because there's an electric field caused by the other point charge uh at its location okay now we got to talk about the direction uh of the electric field as well and it turns out uh with positive and negative charges things get a little bit confusing now we dealt with gravity so there was only one kind of mass it wasn't like there was positive mass and negative mass or mass and anti-mass so and there was only an attractive Force but now because there's now an attractive force and a repulsive force and there's two different types of charge things get a little confusing and we'll find out that in almost every case uh things were defined in the context of talking about a positive charge and they just know that for the negative charge things tend to be the opposite same thing for electric field here so uh if we take a positive charge and we're going to call this an imaginary positive charge and put it in between q1 and Q2 here so and we're going to talk about well what is the direction of the electrostatic force that would be experienced by this imaginary positive charge this imaginary positive charge is more commonly called a positive test charge uh and in this case it would feel a repulsive force from this uh from q1 here which would point to the right and it would feel an attractive Force to Q2 which would also point to the right and two forces uh pointing to the right adds up to a bigger force that points to the right and so the force it would feel points to the right well it turns out that the electric field always points in the direction the same direction as the force felt by a positive test charge so not only does the force felt by this test charge this positive test charge point to the right but the electric field points to the right now what's tricky about this is whe whether we have a charge at that location or not there's an electric field that's going to be defined and it's going to point to the right because what if instead we actually didn't have a test charge but had an actual negative charge at that location well again the direction of the electric field is defined as if there were a positive charge there and it would still point to the right it points in the same direction as a force experienced by a positive test charge whether or not there's a positive charge there or uh or not in this case the force experienced by that negative charge is going to point to the left it's attractive to q1 repulsive away from Q2 and the force is going to point to the left but the electric field by definition is still going to point to the right so be careful with that so again the the direction of the electric field is always in the same Direction uh as the force felt by an imaginary positive test charge whether you actually have an imaginary positive test charge or not all right this makes a good transition to talking about what we call electric field lines and electric field lines are a great way to kind of graphically uh give some indications about the electric field in certain areas so couple rules about electric field lines so electric field lines usually originate from positive charges and terminate on negative charges so now it turns out that sometimes uh you know if there's no balance of charge in in your diagram you might have multiple positive and negative charges well if there's excess positive or negative charge you may have some of the field lines originating uh infinitely far away not actually originating on the diagram they're coming from outside the diagram or you might have them terminating infinitely far away and kind of just exiting the diagram but if the Char is are balanced then every field line should either start from a positive charge or end at a negative charge never the other way around now there's also a quantitative nature to these electric field lines as well so these electric field lines we draw as arrows so the number of arrows corresponds in some way shape or form to the magnitude now it's not like you know there's a number value assigned to every arrow what there is though is a relationship so if these two lovely point charges were on the same diagram then I could come to the conclusion that they have the same magnitude of charge now I might not know what the charge is if if it's not given but I can conclude the same magnitude because the number of field lines originating from this positive charge is exactly the same number of field lines that's terminating on this negative charge and that's how I would conclude they have the same exact magnitude now if the number of field lines was not the same which everyone had less would be a charge of lower magnitude which everyone had more would be a charge of Greater magnitude also uh because if you have uh greater positive charge or greater negative charge you're going to have more field lines around it so and then you're also going to have more field lines that are going to be closer together as a result too and so it turns out in general the closer field lines are together that kind of symbolizes a stronger electric field at that point in space as well so just a couple things to keep in mind for properly drawing electric field lines uh on the study guide and I'll put it up on the screen here there's also a picture of an electric dipole that kind of shows if these two might be part of the same diagram so uh fixed in space but positive on one side negative on the other side and you'd find that electric field lines would start at the positive charge terminate on the negative and they wouldn't just go in a straight line though you'd have them curving in space from one to the other and stuff as well and the further you get away the further the field lines would get away as they make their way around symbolizing that the further you get away from these charges the lower the value of the electric field there as well all right so from here we're ready to do a little bit of pluggin and chugging first question we're going to do here says what is the electrostatic force acting on a 1.0 micrum Point charge in a region where there is a constant electric field of magnitude 8.0 Newtons per K we'll start there well again we got the magnitude of the charge we've got the magnitude of the electric field and it's simply Q * e to get that electrostatic force and so in this case Q * e is just going to work out to give us 1 * 10 -6 K * 8.0 Newtons per Kum and we can see that this is going to work out to give us 8.0 * 10 -6 Newtons the kums will cancel okay so there's the force second half of the equation though says if the mass of the point charge is 0.0020 G what is its resulting acceleration and so in this case we know the net force acting on it we know its mass now we want to find it acceleration and that's just Newton's Second Law here in this case we'll use f equals Ma and if we rearrange the solve for acceleration it's equal to force over Mass which in this case was that force of 8 * 10 -6 Newtons and here the mass is given is 0.0020 G that's 2 * 103 G but again this has got to be in kilogram SI units and that's going to be 2 * 10 -6 kilog in this case we can see that this is going to come out to exactly we want two sigfigs but 4.0 m/ second squared for that acceleration so again a force of 8.0 * 10- 6 Newtons an acceleration of 4.0 m/s squ so the last question here comes with this lovely diagram and question says for the following arrangement of point charges at what location along the horizontal axis is the electric field zero so at some point here so we know electric field due to point charges is KQ over R2 and at some point along this Axis or at least in one point anyways uh somewhere we're saying that it has to be zero and there's two things contributing at two different point charges but at some locations going to be zero and we got really three different regions to consider left of both point charges in between the two point charges or somewhere to the right of both point charges and we first want to take a qualitative look at those three different regions uh so maybe we can rule some of them out and the first one I want to take a look at is in between not because there's any Rhyme or Reason Just because I know it's going to be one of the ones we rule out pretty quickly now if we take an Envision having a positive test charge right in the middle there so the idea is this we said that a the electric the direction of an electric field is the same as what uh the force felt by a positive test charge if we can come up with a region where a positive test charge might not feel any force that would be the place where it might not have an electric field as well all right so in this case it's going to feel an attractive Force to the negative charge on the left that's going to be a force pointing to the left it's going to feel a repulsive force from the positive charge on its right so that's going to also point to the left and so these two components to the force that this positive test charge would feel both point to the left and again depending on where in this region you put it no matter what though the the magnitude of of how much these Point F left is is going to change but they both Point left when you add them together you're going to get a bigger resultant Force pointing left no matter what well there's no way there there's a no net force and therefore no uh zero electric field possible there as well now if we consider a location over here on the right so here with the negative charge here there's going to be an attractive Force pointing to the right but a repulsive force in the positive charge over here that would point to the left and if there's a way that these two components might be equal in magnitude since they're opposite direction they can potentially add to zero we can't just rule it out right off the bat same Fashion on the other side Vision a positive test charge here there's going to be repulsive Force felt due to the positive charge and an attractive Force felt to the left due to the negative charge and again they're in opposite directions if it's possible for them to be equal in magnitude these two components but opposite direction then it could be zero now we'll keep in mind a couple other things here so your electric field is proportional to the magnitude of the charge and it's in inversely proportional to the distance of Separation squared now we also want to then consider which of these might be larger based on those considerations so go back and start this process over again I know the directions but I also want to represent the magnitudes Now by bigger and smaller arrows and so here from a purely a charge standpoint which one of these two charges is going to cause a larger electric field well whichever one has a larger charge because electric field is proportional to charge so based on charge considerations alone it's the direction of whatever the positive 9 micrum charge causes for this positive charge over here that's going to be the force pointing to the left that's going to be the larger one based on charge considerations but for distance the smaller the distance the larger the impact the larger the magnitude of the electric field and it's closer to the negative charge which is contributing a component pointing to the right and so all of a sudden there's going to be a possible for these to be equal in magnitude so one's larger based on charge considerations one's going to be a larger component based on distance considerations at some point at some distance to the left of both of these charges those potentially get have the same magnitude and add up to zero on the other side we're going to see that's not going to be possible based on charge considerations again the 9 micrum charge is bigger that's a repulsive the larger force or sorry and larger electric field is going to point to the right it's also the one that's closer to this charge over here anywhere over here and So based on distance considerations the bigger one is also still going to be the one pointing to the right the one to the left is going to be smaller in any case no matter where you stick it over here and there's no way for these two components that point opposite directions to be equal in magnitude no matter how you add them you're going to get a net force pointing to the right regardless of where you envision uh the location to the right of both these charges so this is going to be the one that we have to consider the other two we've kind of ruled out and now we've got a little bit of math to do so and we to do this the easy way and the hard way it turns out I picked some really nice numbers here you can kind of re reason this out but much of the time you can kind of ballpark this out a little bit now we said that the electric field is proportional to Q so and again somewhere over here based on charge considerations alone we know that the 9 micrum charge is going to have a larger impact on electric field its component is going to be larger how many times larger nine times larger because nine microc is nine times bigger than one micrum so based on the fact that electric field is proportional to q nine times bigger based on on the 9 micrum charge okay now which one's going to be larger based on the distance of Separation well it's going to be this one and I need R so I need r s here to be nine times smaller for the negative charge so again there's an inverse relationship so if I want the electric field to be larger for the one micrum charge here so which is pointing the opposite direction so I want to be closer well it is closer I need it to be a certain amount closer I need r s to be 9 * closer and if R 2 is 9 times closer that means R take the square root is three times closer and so we can take a look at this distance right here which I'll call X and compare it to this distance right here which is 10 mm plus X and ultimately I need X to be three times closer than 10 + x and you could write that as an equation you could say that x = 13 of 10 + x and then solve for x and so in this case I'll both sides by 3 and I'll get 3x = 10 + x I'll subtract X off and we'll get 2x = 5 and therefore x uh how about uh 2x equal 10 still let's get that right divide by two and you get x equal to 5 and all of a sudden we see that X is going to need to be 5 millim left of the negative charge and that way this distance right here would be 15 millimeters and that's exactly three times bigger five millimeters compared to 15 millimeters exactly the relationship we said we needed now if the numbers weren't nice you could still kind of ballpark this in your head and that way when you calculate the answer it might make sense but how would you actually go about solving this well what you'd ultimately do is you'd say k q1 over r1^ 2+ K Q2 over R22 it's going to add up to zero this is what you'd ultimately do I'm going to leave the K's in there for a minute because they're going to cancel anyways in fact I could cancel them out right now could the whole equation both sides by K and it would fall out but I'll leave them in there for now uh K * q1 I make q1 this 1 micro charge so 1.0 * 10 -6 km and the distance of Separation here we called X so and in this case uh the force felt by a positive charge is going to be to the right so the electric field is going to be to the right and I'm going to call that the positive X Direction so I'm make this term positive and then for Q2 here I'm going to make that the 9 micrum charge and we're going to have K * 9 * 10 -6 K all over R2 2 which which is going to be 10 + x^ 2 and in this case relative to the positive charge over here it's going to be a repulsive Force pointing the left which means there's an electric field pointing to the left I'm going to make that negative and these are going to add up to zero all right the algebra here is a little worse than the equation we solved here but ultimately reduces down to the same thing as we'll see and so in this case we rearrange this just a little bit so K * 1.0 * 106 I'm going to leave off the units to make this a little bit easier all over x^2 so I'm going to add this to the other side so K * 9 * 10 -6 all over 10 + x^ SAR and a couple things we can see one I can definitely divide out those KS but also I can divide out these 10 of the sixes to make my life a little bit easier as well in this case this reduces down to 1 over x^2 = 9 over 10 + x^ SAR you got a couple couple different paths you can choose here so you can do some cross multiplying and you could foil this 10 + x quantity squar out and stuff like that or you might realize that you've got perfect squares on both sides the square root of one is one square of x square is x square Ro of 9 is 3 S of 10 plus xty squ is just 10 plus X because you got perfect squares in both the numerator and denominator on both sides you can just take a square root and make your life a lot easier and so in this case we're going to get 1X = 3/ 10 + X and now I'll cross multiply we'll get 3x = 10 + x I'll subtract off X and I'm going to get 2x = 10 and once again x = 5 just like we saw here but ultimately The Final Answer could be phrased in a couple different ways we could say that the location where the electric field is zero is at a point in space located 5 mm left of the ne - 1 micrum charge or it's at a location in space 15 mm to the left of the positive 9 micrum charge either one of those obviously correct if you found this lesson helpful consider giving it a like happy studying