hello class professor anderson here let's see if we can calculate the e field of a point charge and let's do it using gauss's law so we know what a point charge looks like here's our point charge let's call it positive q and that point charge has an e field that is pointing everywhere radially outward and you'll see why i'm going to draw it with dashed lines in just a second okay so there's the e field around my point charge plus q how do we calculate this e field using gauss's law what is the approach well the first thing we do is we write down gauss's law so we know exactly what gauss's law is gauss's law is the following integral e dot d a equals q enclosed over epsilon naught once once you write down gauss's law you need to draw a surface so let's draw a surface around this thing and since we have spherical symmetry let's draw a sphere and now you see why i drew it with a dashed line because the e fields are inside until they poke out through the surface so this is what i'm trying to represent here the fields are inside this ball and then they poke out through the surface the left side of this thing is the total flux right this is the flux coming out of our closed surface as gao said it's proportional to the enclosed charge so one thing that we can do right away is let's put a radius on this sphere so we're going to say that this radius is r okay you can draw any size sphere you want call it r in radius we also need to think about d a what is d a in this case d a is a surface area element of the sphere so if i think about the d a it's a little slice of surface area that looks like that okay but an area has a particular direction to it and that direction is always perpendicular to the plane of that surface so if you have a little piece of this sphere d a is going out which is good because we know that e is also pointing out right everywhere e field is pointing radially outward we can say e has a radial component that's exactly parallel to d a and that's going to simplify this quite a bit the other thing that we can say is e anywhere around on this sphere has to have the exact same magnitude why because it's always r away from the point charge and by symmetry if you flip the whole problem over you have to get the same result right you have to get an e field that has the exact same strength over here as it would over on the other side okay so what does this equation now become well it becomes quite simply this integral of e dot d a becomes e times d a times cosine of the angle between them all of that is equal to q enclosed over epsilon naught but we just said that e is a constant in magnitude around the whole thing and so e can come right out of the integral e integral va we also said that phi is the angle between e and d a and that is 0 degrees right e is this way d a is the same way they're parallel all right this is looking good it's going to simplify quite a bit what does this whole left side become well cosine of theta if cosine of phi if it's zero that thing becomes one and so we're just left with that over on the right side q enclosed is just pos plot sorry positive q right it's just one charge that's in there positive q and so this becomes q over epsilon naught and now look at this right here integral of d a closed surface integral what does that mean that is the area of the sphere the surface area of the sphere right pretty cool because that whole thing just becomes 4 pi r squared that's the surface area of a sphere equals q over epsilon naught and now we are basically done we just have to divide and we get e is equal to 1 over 4 pi epsilon not times q over r squared and that's the electric field of a pooled charge now you look at this answer and you say okay that looks right except i don't have any direction there where do we get our direction from the direction comes back from the symmetry in the problem when we first wrote this down we said the symmetry of the problem means the electric field has to be pointing radial it's either radially out if it's a positive charge it's radially in if it's a negative charge in this case we had a positive charge and so we can get to our final result you say e is 1 over 4 pi epsilon not q over r squared r hat okay electric field of a point charge calculated using gauss's law all right hopefully that's clear cheers