Electric Flux and Gauss’s Law | Electronics Basics #6

hello dan here from howtomechatronics.com in this video we will learn about electric flux and gauss's law in order to understand gauss's law first we need to understand the term electric flux electric flux is the rate of flow of electric field through a given surface it is the amount of electric field penetrating a surface and that surface can be open or closed first we will take a look at an example of electric flux through an open surface these red lines represent a uniform electric field we will bring in that field a rectangle which is an open

area and we will divide the area into very small elements each with size d a the a is called a differential of area now we are going to make the area d a a vector with a magnitude d a the vector direction is always perpendicular to the small element d a the electric flux that passes through this small area d phi also called a differential of flux is defined as a dot product of the magnitude of the electric field e and the magnitude of the vector area d a times the angle between these two vectors

theta the total flux is going to be the integral of d phi or the integral over the entire area of e dot d a it is a scalar quantity and the end result can be positive for negative if the flux is going from the inside to the outside we can call that a positive flux and if it's going from the outside to the inside that's a negative flux the unit of electric flux is newton meters squared per coulomb to get a better understanding of what electric flux is i will bring into this electric field three

rectangles in fact these rectangles represent one rectangle with different orientations now let's explain the flux through each one of those open areas in the first case the area is perpendicular to the electric field and the angle between their vectors theta is zero cosine of zero is one so the electric flux is going to be e times d a here we have the maximum flux in the second case the angle between e and d a theta is 60 degrees and cosine of 60 degrees is 0.5 so the electric flux will be half of e times d

a in the third case the area is parallel to the electric field which means that their vectors are perpendicular to each other and the angle theta between them is 90 degrees cosine of 90 degrees is zero so the electric flux here will be zero this means that nothing goes through this rectangle so the flux is zero now let's take a look at a surface that is completely closed how do we define flux here we can put some normals the a's in different directions by convention the normal to the closed surface always points from the inside

to the outside now we can calculate the total flux going through this closed surface the total flux is equal to the integral of d phi over the entire surface which we write as the integral over the closed surface of e dot d a the total flux can be positive negative or equal to zero if the same amount of flux is entering and leaving the surface we have zero total flux if more flux is leaving than entering the surface then the total flux is positive opposite if more flux is entering than leaving the surface we have

negative total flux let's take a look at another example and see how the electric flux is related to gauss's law we have a point charge plus q in the center of a sphere with radius r now we will take a small segment d a which vector is perpendicular to the surface and is radially outward the electric field generated by q at this point is also radially outward this means that d a and e anywhere on the surface of this sphere are parallel to each other and the angle between them theta is zero and cosine of

zero is one the differential of flux through the small surface area d phi is equal to e d a the total flux phi is going to be the integral of d phi which is the integral over the closed surface eda the magnitude of the electric field everywhere is the same because the distance from the charge is the same at each point so we can pull that out of the integral and we are left with e times a the total area of the sphere is 4 pi r squared and the total flux through this closed surface

is simply e times 4 pi r squared from the previous videos we know that e is equal to k times q divided by r squared which is equal to q divided by 4 pi epsilon not r squared here we can cancel out 4 pi r squared and we can notice that the total flux is equal to q divided by epsilon naught where epsilon naught is the permittivity of free space the flux doesn't depend on the distance r we would get the same result no matter the size of the closed surface around the point charge what

if we bring more charges inside the closed surface the equation should also hold for any system of charges inside this leads us to the gauss's law which says that the electric flux going through a closed surface is the sum of all charges q inside the closed surface divided by permittivity of free space epsilon naught if the flux is zero that means there is no net charge inside the shape there could be positive and negative charges inside the shape but the net is zero no matter how weird the shape gauss's law always holds as long as

there is a perfect symmetry in the charge distribution inside the surface so in order to calculate the electric field you need a symmetry and there are three types of symmetry spherical cylindrical and planar symmetry we will start with the spherical symmetry this is a thin hollow sphere with radius r and we will bring a positive charge q into the thin shell which is uniformly distributed now we need to find the electric field inside the sphere at a distance r1 from the center and outside the sphere at the distance r2 from the center to do that

we need to determine our gaussian surface in this case we will choose concentric spheres as gaussian surfaces one smaller with radius r1 and other larger with radius r2 now we need to use two symmetry arguments that will help us calculate the electric field the first symmetry argument shows that the magnitude of the electric field is the same at any point since the charge here is uniformly distributed the second symmetry argument shows that if there is an electric field it must point either radially outwards or radially inverse in this example we have a positive charge which

means that the field is pointing outwards from the previous equations we know that the surface area of a sphere which is 4 pi r squared times the magnitude of the electric field e is equal to the charge inside the sphere q divided by the permittivity of free space epsilon not however we don't have a charge inside the smaller sphere so the electric field is zero if a closed surface has no net charge enclosed by it then the net flux to it will be zero now let's see what happens with the larger sphere the symmetry arguments

hold for this sphere as well but if we take a look at the equation we will notice that q is not zero because there is a charge inside that sphere so the magnitude of the electric field will be equal to the charge and closed divided by 4 pi epsilon not r2 squared if we draw a graph with the distance on the x-axis and the magnitude of the electric field on the y-axis we can notice the following up to the point r which is the radius of our initial sphere we have no electric field but then

it reaches its maximum value and decreases as the distance increases second type of symmetry is cylindrical symmetry let's say we have an infinite line of positive charge with uniform linear charge density lambda and we want to figure out what the electric field is at some point above the line at distance r here we will choose a cylinder as a gaussian surface with a center along the line of charge we don't have an electric field through the end cups the electric field will be pointing out through the walls of the cylinder also we have symmetry here

which allows us to use the gauss's law in order to calculate the electric field we can calculate the flux using the same equation that we used previously but now we need to find the surface area around the cylinder including the wall without the end caps for that purpose we need to cut the cylinder along its length and we will find out that the area is equal to 2 pi rl so 2 pi rl times e is equal to the charge enclosed divided by epsilon naught the charge density lambda is the total charge q per length

l so the charge enclosed is equal to lambda l so 2 pi r l e is equal to lambda l divided by epsilon naught the electric field is equal to lambda l divided by 2 pi r l epsilon naught l cancels out so the electric field is equal to lambda divided by 2 pi r epsilon naught the last type of symmetry is planar symmetry in this example we have a flat infinite large horizontal plate we'll bring a charge onto this plate with a uniform charge density sigma sigma is actually an amount of charge per area

and is expressed in coulombs per squared meter now we want to calculate the electric field in the surrounding area of this plate let's say at a distance d in this case we are going to choose a cylinder again as a gaussian surface the cylinder intersects the plate and in that intersection we have a charge enclosed in order to be able to calculate the electric field we need to meet three conditions first the cylinder end cups with an area a must be parallel to the plate second the walls of the cylinders must be perpendicular to the

plate third the distance from the plate to the end cup's d must be the same above and below the plate now that we met the symmetry requirements we can calculate the electric field using the gauss's law we are not going to have any horizontal components of the electric field only vertical coming out of the two end cups sigma is equal to the charge divided by the surface and from this equation we can see that the charge q is equal to sigma times the area the flux from the wall of the cylinder is equal to zero

so the total flux consists of two components the flux through the top cup plus the flux through the bottom cup of the cylinder this is equal to q enclosed divided by epsilon naught or sigma a divided by epsilon not but also the flux through the top and the flux through the bottom can be expressed as ea so the total flux is equal to 2 ea finally the electric field is equal to sigma divided by 2 epsilon not if the plate is positively charged the electric field would be pointing outward and if it is negatively charged

the electric field will be pointing inwards if we draw a graph with the distance d on the x-axis and the electric field on the y-axis we can notice that the electric field has a constant value of sigma over 2 epsilon not and it doesn't depend on the distance from the plane now let's take a look at another more complex situation of two infinitely large parallel plates the first plate has a surface charge density plus sigma and the plate below has a surface charge density minus sigma the distance between them is d so what is the

electric field anywhere in the space the positively charged plate has an electric field pointing away from the plate equal to sigma divided by two epsilon naught it doesn't depend on the distance from the plate so it continues below the negatively charged plate has an electric field pointing towards the plate also equal to sigma divided by 2 epsilon naught in order to calculate the total electric field we are going to use the superposition principle by adding vectors the vectors that are in the opposite direction cancel out so the electric field there is zero the vectors between

the plates are in the same direction so the electric field is sigma divided by epsilon naught here's how the electric field lines would look like they will be pointing away from the positively charged plate and towards the negatively charged plate and the electric field outside will be zero okay so that would be all for gauss's law i hope this video was helpful and you learned something new don't forget to subscribe and for more tutorials visit my website howtomechatronics.com