Transcript for:
Electric Fields and Forces

let's revise the whole of AQA electric Fields we're going to start off with forces between charges in a vacuum imagine that I have a couple of charges so let's say that this one here is q1 and let's have another one which is Q2 now those two charges are a distance R from one another depending on whether they're positive or negative they will either repel or they're going to attract let's say that they're both positive for the sake of argument there will be a repulsive force between them this means that the charge Q2 will experience a force this way and the same equal and opposite force will also be present on the charge q1 the magnitude of this force will be given by a constant which is conveniently written as 1 over 4 Pi Epsilon and then we have the magnitude of the two charges q1 Q2 divided by the square of the distance between them now e not over here is known as the permitivity of free space its value is given in the formula book booklet it's around 8.85 * 10 ^ of -12 something important to note is is that if one of these charges is suddenly negative then this force will be attractive depending on the charges another thing we need to mention is that this is called the permitivity of free space because this is the equation that applies in a vacuum however air can be treated as a vacuum so I'm just going to write down over here that the air can be treated as a vacuum one final thing to consider is that if we have some charged uh spheres the charge may be considered to act at the center of that sphere and if we ever need to figure out the distance the distance R will be this distance over here let's do an example problem we have two charge spheres and they're shown over here and they exert an electrostatic force of attraction on each other we decrease the distance between their centers by factor of two what will the new Force F2 be in terms of the original Force F okay so the equation I'm going to do this I part of this problem uh over here is that the new Force F2 is simply going to be the product of the two charges q1 and Q2 divided by some constants 4 Pi Epsilon R 2 now if R is reduced by a factor of 2 then the denominator of the fraction will decrease by a factor of four meaning that the new force will increase by a factor of four because of that F2 will be four times the original Force part two and that's new dist each sphere is given an additional charge minus Q okay so this means that the charge on this one is going to be + 4 q because 5 q take away the new charge minus Q is 4 q but then this one here will turn to Min - 3 q and then minus an additional one will be minus 4 q find the new Force yet again F3 in terms of the original Force F the might be a bit more algebra on this one so uh let's write down the new Force F3 which is going to be the product of the two charges which is going to be given by 4 q multiply the magnitude of the other charge which is also 4 q over 4 Pi Epsilon uh R over 2 2ar um it's r/2 because because we are at that new distance which was half the original distance R so F3 we're getting somewhere we have an expression for it is going to be 16 Q over 4 piun Epsilon r/ 2 squared okay so how does that compare with the original Force F but we also know that F2 was let's do that in a different color was actually before equal to 5 q * that by 3 Q over 4 Pi Epson r/ 2^ 2 and this was equal to 4 times the original Force now I'm going to pick a common variable between them this appears quite a lot in AQA physics so this could be just a charge Q for instance 5 * 3 o there's a missing Square here really important let's add that um 5 * 3 will be equal to 15 q^ 2 over 4 piun r/ 2^ 2 will be equal to 4 F and now I have a common variable between those two expressions which is q^2 so should we just rearrange for q^2 which is going to be equal to 4 uh over 15 and then I'm going to multiply by 4 piun Epsilon r/ 2^ 2 and then multiply that by F okay now let's plug this into here and we should get an expression in terms of f so if we were to do that we're going to get that F3 will be equal to 16 over 4 piun Epsilon r/ 2^ 2 now the 16 will then be multiplied by Q ^2 and run than Q ^2 I am going to write * 4 ided 15 and then multiplied by 4 Pi Epsilon R over 2 sared multiply this by f look at this almost all of this is going to cancel out so the 4 Pi Epsilon over 2^ 2 is going to cancel out and all that we're left with is 16 * 4 / 15 so what does that give us if we were to put this into a calculator as a decimal it gives us around 4.26 uh F which is around let's call it 4.3 multip by F let's do another practice problem A variation of this problem I have seen appear in multiple different forms so we have a positively charged charged sphere of a certain Mass .25 kg is suspended on an insulated piece of string next to a negatively charged wall okay so this means that those things are the wall and the sphere are going to attract meaning that the sphere will get will get tilted from the equilibrium position to another equilibrium position over here on the left find the electrical force of attraction between the sphere and the wall it's always worth resolving the force to see what's actually happening in this problem now first of all there's going to be a downward force due to gravity which is the weight and this here will be acting pretty much straight down let's call that mg now in order for the sphere to be in equilibrium the upwards force of tension will have to be equal to that so the tension can be resolved into two components let's call this one I don't know let's call that t this one here could be um T cosine of theta so we can say that the adjacent component T cos Theta is equal to mg there's also going to be a force on the sphere an attractive Force let's pick a different color for that one which is going to be in this direction and and that's simply the electrical force the horizontal component of the tension will act in the opposite direction and will have to be exactly equal to the other one let's draw it all the way to the end it's not really drawn to scale but this blue arrow has to be equal to that Arrow over here so T sign of this angle here will have to be equal to the electrical force Fe even though we can solve this problem quicker without drawing this triangle A variation of this problem May potentially appear so I think it's really worth explaining how it works so because T sin Theta is equal to the electrical force all we need to figure out is essentially our opposite component we already have our adjacent component because that's just well equal to mg in a way we actually have a triangle which is just a 90° triangle with the vertical component being equal to mg and the horizontal component being equal to the electrical force we have the angle here so we can directly say that the tangent of 40° is equal to the opposite over the adjacent So Tan of 40 will then be equal to the electrical force I've just call that Fe divid by the adjacent which is mg meaning that the electrical force will just be equal to mg multipli this by the tangent of 40° now our mass in this problem was .25 kg multiply this by 9.81 multiply this by the tangent of 4 ° and putting this into a calculator we're going to get around 2 uh 057 Newtons uh shall we just call it 2.1 2.1 Newtons and now let's compare the gravitational and the electrostatic force between subatomic particles we need to find the ratio of the electrostatic repulsion to the gravitational attractive Force so the easiest way is just to draw a little line because remember ratio is just division so at the top we're going to have the electrostatic repulsive Force I'm just going to call that Fe at the bottom I'm just going to write F subscript G for the gravitational force so my repulsive force is going to be given by the magnitude of the two charges they're going to be the same but let's just write q1 Q2 and then we're going to be dividing that by 4 Pi Epsilon r² then at the bottom if you remember the law for gravitational attraction is equal to M1 M2 multiply by the gravitational constant G divide that by the distance between them which is R 2 the factor of the distance 1/ R 2 is going to be the same for both of these so that can be cancelled out okay let's plug in some numbers the charges and the masses are going to be the same so just Den denominator of this fraction is going to be q1 Q2 which is 1.6 * 10 ^ -19 this is the charge of a proton and it's one multiplied by the other so that's why I'm going to square it then we're going to be dividing that by 4 piun * by Epsilon which is 8 .85 * 10 ^ of -12 so this here is just the denominator of our fraction then we need to divide by gm1 M2 so it's because it's two protons it's just going to be one multiply by 1 over the gravitational constant G is 6.67 * 10 ^ -1 then the mass of a proton if memory serves me right it's around 1.67 * 10^ of -27 there's two of them so we need to square it plugging this into a calculator we're actually going to get an enormous number somewhere around 1. 237 that's called 1.2 multiplied by 10 ^ of 36 the electrical force is orders of magnitude larger than the gravitational force now going to have a look at representing electric Fields by electric field lines now first of all what is an electric field well it's simply a region of space uh if we were to place a small positive test charge for instance uh if I was to place right over here a small little proton like a small positive test charge next to this positive charge it will experience an electric force the electric force along here will actually be in this exact Direction because it's repulsive it will be to the right now the electric field lines show that direction they show the direction of the force SL acceleration on a small positive charge that's been placed in the field I'm just going to underline the word positive because that's really important that's how we Define the whole of electric fields in terms of positive test charges drawing the electric field lines is very very similar to gravitational field field lines however they can be positive or negative because the force can be both attractive or repulsive if the charge is negative the field lines go into the negative charge if it's positive they come out of the positive charge because of that we can also say that the electric field lines always go from positive to negative what are some other rules for their drawing well they enter and leave surfaces at 90° the density of the field lines means a higher electric field strength for instance if I was to have a charge let's say plus something but the field lines were so much denser then this would mean that the electric field strength at that around that charge is higher meaning that probably there is more charge right over here if I had a positive charge next to a negative charge we can just follow all those rules first of all between them the electric field will be going from positive to negative and following this rule that they enter and leave surface at 90° the field line is going to leave the positive charge it's going to curve away then it's going to enter at 90° and right over here following this one it's going to leave at 90° curve away and then enter always from positive to negative at each point if they were both positive the field lines will end up canceling between them why is that because there will be a field line coming from here in this Direction one from there in this direction and they're both in the same direction so they end up canceling each other out another type of question that appears is a charge sphere next to a wall we just follow the rules field lines go from positive to negative they leave at 90° and they enter at 90° at pretty much all the points and now let's talk about electric field strength this here is a really important definition electric field strength is defined as the force per unit charge on a small positive charge this is typically a two marker with one Mark going for the force per unit charge and one that it acts on a small positive charge there are several equations which uh we can express this definition with the first important one and that is the main um electric field equation is that electric field strength is equal to the amount of force divided by the charge now in this equation Q is the charge Which experiences the force another really important equation is for radial Fields since the electric field strength is equal to F / Q um if we remember the electrical force is equal to q1 Q2 over 4 Pi Epsilon r² and then we need to divide by Q so we doing 1 / Q where Q is the force that is the charge that experiences the force now these are going to cancel and what we're left with in this case is just Q2 over 4 piun Epsilon R 2 now Q2 is in this case the charge which creates the field or the charge which is the source of the field the index is actually typically removed so people tend to in general just write that the electric field is given by Q over 4 Pi Epsilon r² this only applies for a point charge or for a spherical charge essentially for radial Fields only on the other hand e is equal to F over Q can be applied for any electric field out there there are a couple of different units that we can use for electric field strength for now because electric field strength is is equal to force per unit charge force is measured in Newtons charge in kums so the unit we could use is Newtons per Kon you also see later on because of a different equation in uniform electric fields we could also use volt per meter now what about the base units well because Force per unit charge can also be expressed as masstimes acceleration divided by charge which is just current multiply by time we could say well the units of mass are kilogram the units of acceleration are m/s squared the amp is a base unit and the second is also a base unit so what we're going to get is kilog M s^ minus 2 / s is going to give me S to^ of minus 3 and we also have a to^ of minus1 which is the base unit of electric field strength another example problem on electric Fields we've got two charge spheres and they're 1.2 M apart well their centers are 1.2 M apart find the distance D um at which the electric field between the two charges is zero and these is measured from the 2.36 Kum charge so let's just say for the sake of argument that this is the distance here let's draw it in a different color I this one here here will be D so at this point there going to be two electric fields and they should be equal and opposite in practice one of the electric Fields will be from the 2.36 Kum charge which is positive and then there's going to be another one from the 5.73 charge which is also going to be equal and opposite and the two will have to be equal in magnitude so can call one of them I don't know E1 the other one E2 we can say that E1 is equal to E2 because we're dealing with radial fields we can say that q1 over 4 Pi Epsilon R1 s now what will the distance actually be so the distance from the first one will just be D so we can say d^ 2 will be equal to Q2 over 4 piun Epsilon now the distance from one of them to the center of the other is going to be 1.2 takeway D so we're going to have 1.2 take away D and then all of this is squared okay without even plug in the numbers I'm going to do a bit of cancellation so 4 Pi Epsilon note can go and what we want to do is isolate the distances on one side okay so let's just do a bit of rearranging and see what we get um I'm going to say that q1 multiplied by 1.2 take away the all of it squared is going to be equal to Q2 * d^2 this looks looks like it's going to be a quadratic so should we just do expand the bracket so 1.2 minus d^2 what is that going to give me that's going to give me 1.2 2us 2 * 1.2 which is 2.4 multiply this by D and then plus a factor of d^2 and this here will be equal to Q2 * by d^ 2 uh now I can probably plug in the numbers because q1 is that and Q2 is 5.73 and let set up a quadratic so we're going to get 2.36 multiply this by 1.2 s take away 2.4 D + d s will be equal to Q2 which is equal to 5.73 multiply this by d s I'm going to set up the quadratic over on this side now we can bring the factor of minus 5.73 d^2 essentially on this side or the factor will become minus that so this here is going to give me - 5.73 + 2.36 this here will be our a d^2 then - 5.66 4D and then + 3.3 984 is equal to 0er and now we can apply our quadratic formula remember the quadratic formula is given a new formula sheet so um our X or indeed um our distance will be given by minus B so what is that going to give me 5664 and then plus or minus the square root of b^ 2 which is 5.66 4 SAR take away 4 a so plus uh because it's - A so 4 * a which is 3.37 multiply this by C which is [Music] 33984 and we're going to divide all of this by 2 a so it's going to be - 2 * 3.37 now one of our Solutions is going to be between the actual spheres so one of them is 0.469 8 so should we just call that .47 M this actually makes sense I should have picked the point closer to the first sphere because the relative electric field strength of the first sphere is a lot smaller than than the uh than the second sphere the second solution D2 is around - 2.1 uh M well this is not between the Spheres so we can actually discount this it's also not a physical solution because the electric field strength of the um 2.36 charge if we go on the other side of the sphere we'll end up changing its direction now we be going to the left rather than the right so we can completely ignore this solution another type of problem that appears quite often is a geometric shape with some charges on it so for instance we're given a circle and we need to figure out the resultant electric field at the center of the circle if the radius is .5 M so first we need to look at symmetry now we've got a charge + five and a charge plus five here the field between them will consist of a field line from this one that's going to be going this way and a field line from the other one that will also be going the same way and those two field lines will be equal and opposite so that means that the plus 5 Kum and a plus five colum charge will end up just cancelling one another so in a way we can just cancel out those two charges if we have + 5 C here and plus 5 C here well the effect will be the same because through the center there's going to be a line coming out this way and there's going to be a line coming out this way so those two will essentially cancel out the effect of one another so the problem is now greatly simplified because all we need to do is figure out the charge from this five Kum uh charge the electric fueld from there and one from the three colum charge this five here almost doesn't look like a five shall we change it that's a little bit better so those two are meant to be diametrically opposite and to go right through the center like this from the five coolum charge there's going to be an electric field should we just call that E1 which is going to be going this way and then from the three Kum charge there's going to be an electric uh field should we draw this one in red maybe that's going to be going the opposite way because electric field come out of positive charges and those two are both positive so now we need to take away E1 and E2 because geometrically they're going in the opposite direction to find the resultant electric field strength so E1 takeway E2 will be given by let's call it q1 over 4 Pi Epsilon uh r² take away Q2 over 4 Pi Epsilon r² now R in this case is going to be the distance from the charge to that point which is just going to be the radius of the circle so this here is just .50 m in both of these occasions okay so q1 is 5 over 4 piun 8.85 * 10 the^ of -12 multiply this by .50 SAR take away Q2 which is 3 over 4 piun 8.85 * 10^ -2 * .50 squar and this here is going to give me a very large electric field which is 7.2 multiply this by 10 the^ of 10 Newtons per coolon now let's have a look at uniform electric Fields first of all what is a uniform electric field well have a look over here we have two parallel plates one positive and one negative the field lines are just going from positive to negative well the field lines are also equidistant meaning that they're the same distance apart meaning that the electric field strength is the same anywhere between those two plates there's a special equation which we only use for uniform electric fields and that is that the electric field is equal to the voltage between the plates divided by the distance between them we also need to be able to derive this from work done moving a charge between the plates how do we actually do this well remember that work done is equal to in general to force multipli by distance when the force is perpendicular to the direction of motion however work done in moving an electric charge is equal to the magnitude of that charge Q if we were to move a charge let's say Q from one plate to the other and then multiply by the change in voltage or the voltage gap between them which is just the voltage between the plates this here will just be equal to force multili by distance we can rearrange this a little bit and we can say that Delta V over D basically I've just brought the voltage onto the left hand side and I'm going to bring the charge from the left hand side onto the right hand side will be equal to F / Q however Force divided by charge or the force per unit charge remember the magnitude of the electric field strength is just equal to force / char charge so this entire left hand side is just the electric field so we know that the voltage Gap divided by the distance is equal to the electric field and this is where the equation that electric field is equal to voltage over distance actually comes from V is just the voltage between the plates in other words the voltage Gap let's do a practice question we have a charge particle that is placed somewhere between two parallel plates which are separated by distance d and have a potential difference of V between them the charge on the particles is q and its mass is M show that the acceleration of the charge particle due to the plates is given by okay because this is a show question let's underline this command word I'm going to try and give quite a lot of steps in order to work out the acceleration I am going to need the force or the net force AC on it typically in those problem we tend to ignore the effect of gravity unless otherwise specified in the problem so we can say that fnet is equal to mass time acceleration but remember the electric field strength is equal to F over Q meaning that f is equal to e * Q so e * Q will be equal to m a meaning that a will be e q/ M remember though that the electric field is equal to V / D so this here is going to give me VQ over DM which is exactly what we have over here and now let's do a problem which does not ignore gravity on purpose so show that the voltage required to keep a particle of mass m and charge Q stationary between two parallel plates with potential difference V and distance D is given by the following expression if you want to keep the particle stationary we are going to need an upwards Force which balances out the weight so let's say the electrical force is just F this is going to have to be exactly equal to mg which is acting downwards so we can say F has to be equal to mg but remember f is just equal to e * * Q which is equal to mg the electric field is V / D * Q is equal to mg V will be equal to mg bringing the D onto the other side we get MGD divide that by the charge q and those seem to be identical now let's have a look at the trajectory of moving charge particles entering a uniform electric field initially at right angles now what on Earth does that mean well if we have two parallel plates one positive and one negative let's say that we have initially an electron so a negative charge particle entry at the moment this electron enters over here it's going to experience a force the initial force will be towards the opposite plate so if this here was an electron there's going to be a force acting on it a constant force given by EQ um which will be acting towards the positive plate because there's a constant force acting then the electron will start curving towards that plate depending on its initial speed it will either get absorbed by the plate or if it has a Fastener velocity it might be just a very slight curve and might just Escape on the other hand if we had a positive particle so let's say something like a proton and the very moment that it enters it's going to be attracted towards the negative plate well it also gets repelled by the positive plate as well in other words it will go along the field line and it will get curved towards the negative plate and once again depending on the initial speed if it's moving really fast this might be just a very very slight curve and it might um Escape but either way it will be a curve towards the opposite plate now let's do a practice problem we have an electron that enters a region of two parallel plates with a potential difference of 3,000 volts so the electron enters perpendicularly to the field that means purely horizontally in this context with initial speed of 1.5 * 10^ of 7 at the midp point between the plates okay we're given the distance between the two plates which is 25 cm so this means that this distance here will end up being 25 cm over 2 and so this one here the length of the plates is 50 cm or 0.5 M does the electron manage to escape the region of the electric field before it collides with the Positive plate we need to support our answer with calculations and we can ignore the effects of gravity the first step in solving these problems would be to work out the acceleration so Remember the force as soon as we enter the electric field is given by EQ in other words ma is equal to EQ the acceleration will be equal to EQ / M the electric field is the potential difference between the plates divided by the distance between the plates and multiply by the charge okay so the voltage between the plates is 3,000 because it's an electron its charge is 1.6 * 10 ^ of -19 the distance between the plates is 25 cm or .25 it's important to use .25 because the electric field is the same anywhere is given by the distance between the two plates which in this case is a constant okay multiply this by the mass of the electron which memory serves me right 9.11 10 ^ of - 31 kg putting this into a calculator we get around 2.1 multiplied by 10^ 15 m/s squared okay so this here is the vertical acceleration the electron will end up being attracted it towards the positive plate along here and question is is it going fast enough to leave the surface well let's figure out how long it actually takes for the electron to strike the um the vertical surface so this distance here will just be given by .125 m which is just half of it so we can work out the time fr to strike the vertical plate just using suat so I can say that in the vertical Direction the vertical displacement let's call it s y will be equal to the initial speed in the vertical Direction multiply by the time plus a half a t^ s now the initially the velocity was purely horizontal so we can essentially get rid of this term we know what the display M will be is just .25 over2 because it was at the midpoint uh a half we know that we've just worked out the acceleration can just literally rearrange for t meaning that t will be the square root of 2 s y That's the vertical displacement divided by the acceleration okay let's plug in some numbers so 2 * 0.125 divided by the acceleration which is very large 2.1 * 10^ 15 and this here is just a very small number 1.09 * 10 ^ of -8 seconds okay so we have the time of flight this essentially projectile motion but rather than gravitation we have a very strong uh electric field which gives us a much higher acceleration typically now that we have the time of flight we can focus on the horizontal Direction it's important to note that because when the electron is here the only forces that act are in the vertical plane so this means that the horizontal component will remain exactly the same so the horizontal speed at all times will be equal to the initial which was 1.5 * 10^ 7 m/s that remains constant so how much distance will it cover in that time well because the speed is constant let's say that it travels a distance X we can say that X will be equal to the initial horizontal speed let's call it ux multiplied by the time so our speed was 1.5 * 10 ^ 7 m/s multip by the time which is 1.09 * 10^ of -8 seconds and we actually get an answer of 01635 let's just call that around 16 cm so the answer is no the electron will collide with the Positive plate only at a distance of 16 cm so the trajectory is going to be quite um aggressive so let's say that 16 cm is somewhere here it's probably going to do something like this and now let's talk about the electric potential the definition of the electric potential is that it's the work done per unit charge to bring a positive charge from Infinity to a point and we can agree that the electric potential is zero at Infinity an alternative definition typically accept in the mass schames is that it's the electric potential energy that a unit positive charge will have at a point there is a distinction between potential difference and the absolute potential so the formula for the absolute potential for a radial field is that the electric potential is going to be given by 1 over 4 Pi Epsilon assuming that we're in a vacuum and um multiplied by the charge divided by the distance if we're looking at potential difference typically we're looking at Delta V which is the change in this quantity if we wanted to move a charge in some sort of an electric field this would require some work done typically and the equation for the work done let's call that Delta W is the Delta Del W will be equal to the charge multiplied by the change of potential now we can actually use that to get a formula which is not actually on the formula sheet but if we take one of those references to be Infinity where the potential is zero uh change in V can just be equal to V so we can say that electrical potential energy let's say we had two charges q1 and Q2 can also be given by q1 Q2 over 4 Pi Epsilon uh multiply this by R uh this is if we're moving one of those charges in the presence of the other one let's apply this to an example problem we've got a potential given between two points one and two how much work does it take to move a 10 mum charge from .1 to. 2 we're just going to use W is is equal to the amount of charge Times by the potential difference so my charge is just 10 mum so 10 * 10^ minus 3 and my potential difference Delta V is just going to be 200 - -50 so that's going to be 200 - - 50 uh which is + 250 so this is 10 * 10^ minus 3 multiply this by 250 which is just going to give me 2.5 Jews in terms of the graphical representations of the potential with distance uh we're going to have a couple of variations so for a positive charge the variation will look something like this it's similar to the equation Y is equal to 1 /x uh why is that because V is equal to the constant 1 4 Pi Epson Q over R Bas you have V being invers proportional to the distance and somewhere at Infinity this TS tending towards zero so on the other hand if it's a negative charge it's going to look something like this once again approaching zero as R TS towards in Infinity we can figure out the electric field strength at any point just by using the gradient of the tangent we could figure out the tangent at this point so something like this mathematically in thinking about units if we take V and we divide it by an infinite decimal R in reality we're actually differentiating but you can think about this as dividing small quantities of one over the other we are actually going to get units of Q which is the potential for which is the charge for piil R which is the potential divid by R again we're going to get q r 2 well remember this here is just the electric field strength mathematically we would write this really importantly as electric field strength is equal to the change of potential um over the change in distance now sometimes I've found in questions that this Delta R could be disguised somehow for instance you could have the variation of the potential difference against height from the ground or something like this and you may have to figure out the electric field strength by finding the gradient of a tangent at a point sometimes you can also come across a graph such as this one you have the electric field strength against against distance well we could do the reverse we can pick two points let's say one here and point to here then we may need to estimate the area of the graph and if that's the case the reverse of it the area is equal to the potential difference area will be equal to the work done per unit charge to move from one point to the other let's cover equip potentials as well so imagine we have some positive charge we just call it plus Q what do the field lines look around that charge well they're going to be radial lines coming out of the charge so equip potentials are lines where the electric potential is the same the potential around a point charge or a sphere is going to be the same around a circle centered around the charge so if I was anywhere along this circle I would be equidistant from the charge and the potential will be the same or if I was anywhere along here the potential would also be the same there are infinitely many equip potential lines important to remember some rules equip potential lines are perpendicular two field lines and no work is required to travel along an equip potential potential line just a little note that the spacing between the equip potential lines represents the relative strength of the electric field so because the strength of the electric field is decreasing as we move away from the charge the next one would be somewhere quite a bit further apart or something like this but centered uh near the surface that will tell me that the field is actually decreasing with distance from the source and let's do another practice problem what is the change of electric potential energy when a proton is moved from position one to position two which is directly below the electric potential energy is going to be the same anywhere along this vertical line well how do we know this imagine if we were to release a proton from anywhere along this line it is still going to strike the other surface with exactly the same speed because it will have acquired exactly the same amount of energy so anywhere along this line we're going to have the same electric potential energy so the answer to the first question is actually zero what about from position two to position three the distance between the points is 15 cm okay well well the work done will be equal to Q * by Delta V the question that we need to find is what is the potential difference between points 3 and two and we know that this distance here is 15 cm okay we're given the distance between the plates is .5 M so the trick with this is to find the electric field which is equal to V / D which is going to be 100 IDE that by .5 I.E this here is going to be 200 Vol per meter so what potential difference are we going to cover in our journey of 15 cm well because V is equal to e multiplied by the distance we are going to take 200 and then we're multiply this by 15 cm and we're going to get around 30 Vol okay now we're ready we know what our Delta V is we can just plug that into here so we can say the W is going to be Q Delta V the charge of a proton is 1.6 * 10 ^ -19 multiply this by 30 which is going to give me 4.8 * Time by 10^ of -8 Jew well done for revising the whole of electric Fields what you should revise next this capacitors and this is precisely why I should click on this video where I go over the entire AQA syllabus on capacitors click over here