this video presents Pisans equation and Laplace's equation for electrostatic fields in this video we'll review several concepts the versus Poisson equation which is an example of a non-homogeneous differential equation and Laplace's equation which is the homogenous version and we'll use the Laplace equation to solve a very simple example at the end so we begin this discussion with Poisson equation and to arrive at this equation we consider Gauss's law so Gauss's law it says the divergence of the electric flux density field D is equal to the volumetric charge density Rho sub V now we know that there is a constitutive relation between the electric flux density field D and electric field E through this relation assuming that we are in free space so if we substitute this into Gauss's law and rearrange things a little bit we arrive at this divergence relation for e we also know that the electric field E is related to the potential V through gradient so we have this additional relation that we can now substitute back in to the relation that we have here for Gauss's law doing some results in this second-order differential equation that we see here where we now need to take the divergence of a gradient this is not as hard as it seems because we already know how to write for example the gradient of V which we can do here so there's a gradient operation on the right and now we want to dot that with another gradient operator or basically take the divergence so I can just put another gradient operator on the left and a dot in the middle and now we need to compute this dot product so we know we compute dot products by taking the X component of each multiplying them together taking the Y component of each multiplying them together and adding it to the result for the X component and finally the same thing for the Zed component so I could we carry out this simple dot product we arrive in Cartesian coordinates at this relation for divergence of gradient of V and we usually abbreviate that with this del squared and del squared is called the Laplace the operator and we're going to be using it a fair amount to solving these problems so now we have the second-order non-homogeneous differential equation it's not homogeneous because the right-hand side here has nonzero so it's a function of the charge distribution now it's important to realize that the laplacian operator looks different in the various coordinate systems so here is laplacian in cartesian coordinates we just derived that by successively applying divergence and then gradient but in cylindrical coordinates it has a different form slightly more complex form and in spherical coordinates it's even longer so be aware of the fact that the laplacian operator appears differently in the various coordinate systems and much like divergence and gradient and curl do ok so here's an example of a problem that we would like to solve we have a volumetric charge density contained within a volume V and volumetric charge density is Rho so V as usual and that's some point P we are interested in finding some field quantities such as the voltage at point P or the electric field at Point P so from what we've just seen we know that Laplace's equation applies and if we can solve this equation given this volumetric charge density Rho sub V we can find V and from V we know that we can then find e by going backwards so this is a challenging problem to solve or more challenging to solve because it is again a non-homogeneous differential equation and it would be easier to solve if the right-hand side were zero in fact the right-hand side is zero and many parts of this problem if I take this field point or this observation point and I move it outside V so let's say I move it outside of this volume to this point shown here then actually at that point there is no volumetric charge density so the right-hand side is zero and we get the homogeneous equation that we're looking for so that equation where the right-hand side is 0 is called Laplace's equation so it is a special case of Hasan's equation and it holds everywhere outside the source region so everywhere outside this volume v our problem reduces to solving this homogeneous differential equation and it basically turns into a boundary value problem it's important to also realize that we can apply the same concept to many other charge distributions not just volumetric charge distributions so the same is true in turning Hassan's equation until a places equation if we had a surface charge density or a line charge density or even a discrete distribution of point charges so long as our observation is not coincident with the charge density itself or the point charges in the case of a discrete distribution we can turn the problem into solving Laplace's equation and even with a combination of volumetric surface line and discrete charge distributions we can apply the same concept so let's apply this to a simple problem of a parallel plate capacitor we have studied a problem like this before in the context of knowing the charge density on these plates but we're going to approach this a little differently now so let's say that we have two charged plates one at y equals a which has a voltage V of a on it and that voltage is known and another plate at y equals B again with some known voltage V of B on it and our question is what is the voltage and corresponding electric field in this region so we would like to know for example what is the voltage in this area here okay so we begin by having a look at Laplace's equation so Laplace equation applies everywhere here where we are not on one of these two equipotential lines right because we know that there must be a charge distribution on y equals a and the charge distribution on y equals B so so long as we're not focused on what what is the voltage on those points and in fact we know what the voltage is already on those points away from those points we are trying to solve Laplace's equation now we can clearly work in Cartesian coordinates for this problem so if we expand the laplacian operator we know that we have the second derivative of V with respect to X plus the second derivative of V with respect to Y plus the second derivative of V with respect to Z and that must be equal to zero but in this problem there's only variation in the Y direction so we don't need to worry about these two derivatives here so now we can integrate both sides so if I integrate both sides once then I would just get the first order derivative of voltage with respect to Y and the right-hand side must turn into a constant let's call that constant a and then if I differentiate again to get just plain V then I should get a times y plus another constant B so the question or problem becomes now finding these two unknown constants a and B but we know the voltage at these two positions maybe will simplify things a little bit and call this V sub a and this upper voltage here V sub B so let's now substitute in various points so let's evaluate the voltage first of all at y equals a well we know that voltage is equal to V sub a and this would be equal to a times lowercase a plus B and so we have now one equation and we have two unknowns to solve for but we also know that if we evaluate the voltage at y equals B this is also equal to VB and this would be equal to a times lowercase B plus B so now we have two equations and two unknowns and we can proceed to solve these equations for these constants a and B to simplify our problem a bit let's consider the fact that we actually only need to know the voltage difference between these two plates so I can reassign the voltage on the bottom plate to be a reference which means that it's zero volts and we'll say that the top plate is charged to a voltage l will be not with respect to the bottom plate so then if I take my previous equations and for example evaluate V at a we know that this is now zero and equal to our previous equation and so this allows us to solve for one of our constants or propose at least an expression for one of our constants let's say B then if we look at the upper plate we know that this is also charged unknown voltage here and we have a separate expression for that but now we can substitute in our previous relation for B and say that this is just a times B minus a times a or if you want a times B minus a and this is equal to V naught so now we can solve for one of our constants a this constant a is equal to V naught divided by B minus a which is the distance between the two plates we can then substitute this back into our expression for B which we have up above so this is just going to be V naught over B minus a but the negative of that because there was a minus sign out in front so now we have a expression for voltage as a function of position Y we substitute in our constant a which is V naught over B minus a times y and then we substitute our constant B which is V naught over B minus a so now we have a complete expression for these and if I wanted we could have replace these terms in the denominator with let's say some value D where D is the plate separation between these two surfaces so now we have a complete expression for the voltage that we derived using Laplace equation we can proceed to find the electric field between these plates because we know that there's a gradient relation between potential and electric field so let's do that next so we know that electric field is minus the gradient of our potential field and we have an expression for the potential field first of all we realize that there's only going to be a y component of our gradient so we can replace our gradient with this expression here and if I differentiate the previous expression that we derived for potential we're just going to arrive at minus V naught over D times y hat so our electric field is constant between the two plates and it points in the negative y direction or downward so we can draw our electric field that's pointing downward like this in between the two plates and again it is uniform and does not vary with position and we have seen this result before when we considered two parallel charged plates next to each other so in conclusion Hasan's equation reduces to Laplace's equation when we are in the so-called source free region away from the sources which in this case is charges for electrostatic problems this simplifies the solving for the potential field or the electric field in the source free region and it basically gives us another way to solve for potential and a field and electrostatic problems so we have one path where we use Coulomb's law or Gauss's law to find the electric field and integrate that to find voltage here's another way to solve problems like that if you know the voltages already you solve this second-order homogeneous differential equation and then use the solution for B to find other quantities such as the electric field