Gauss Law Made Easy | Electric Flux and Proof Explained

Añadido: 2026-06-02

[00:00:15]Hello, welcome to this video. In this session, we will understand Goss law of electrostatics. an important law that relates electric flux through a closed surface to the change enclosed within it.

[00:00:31]Myself Dr. P Rashar, professor and director IQAC School of Sciences and Humanities, SR University, Vangal, Telangana, India.

[00:00:45]Let's look at the learning outcomes of this topic.

[00:00:52]obtain the Goss law of electrostatics and to apply the Goss law to various types of charge distributions.

[00:01:06]Before we go into the Goss law of electrostatics, let's first have a reook at the Kulum's law.

[00:01:14]Suppose let us say there are two charges Q1 and Q2.

[00:01:18]Let they be positive or negative charges.

[00:01:22]As you know, like charges repel, unlike charges attract.

[00:01:28]So there is a force of repulsion or force of attraction between the two charges depending upon what type of charge they are.

[00:01:37]So according to Kulum's law, the magnitude of electrostatic force between these two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the two.

[00:01:54]So the very famous inverse square law as the distance increases the force of attraction or repulsion decreases. or the charges come too close to each other the separation decreases and the force increases.

[00:02:13]So essentially says that it is directly proportional to the product of the magnitude of charges and inversely proportional to the square. So if you remove the proportionality we get a constant K which is equals to 1x 4 pi epsilon where epsilon is the permitivity of the medium in which these charges are placed.

[00:02:40]If it is free space we will write it as epsilon kn 8.85 into 10^ -12 farad per meter.

[00:02:50]That's a constant.

[00:02:54]So if you put uh the value of k in this expression then the expression lead to 1x 4 pi epsilon kn q1 q2 by square essentially through this we can calculate the force of attraction or repulsion.

[00:03:13]The formula which I have considered is only by considering the scalar nature otherwise the force of attraction or repulsion is directional and we need to consider the vectoral representation but the sake of understanding I've just considered the only the magnitude of the force of attraction or repulsion.

[00:03:37]So with the help of the Kulum's law one can determine the force of attraction or repulsion. Through this we can also determine the electric field intensity experienced by one charge due to the presence of other charge. So we can write E is equals to F by Q.

[00:04:03]Once we determine the force, it is easy to determine the electric field intensity.

[00:04:11]We define the electric field intensity as the force experienced by a unit positive charge. It means that let us say the Q2 charge has a magnitude plus one mean it is a unit positive charge and this Q1 is let us say positive charge and with five kum.

[00:04:32]Okay. So within the influence of 5 kumm positive charge means electric lines of force always radiate outward. So when you keep a unit positive charge definitely the unit positive charge will get repelled with for what force it gets repelled is essentially the electric field intensity that can be calculated using this simple formula.

[00:04:58]Now this uh kulum's law can be applied to only point charges and that to static charges charges of point dimension are termed as a point charges.

[00:05:14]That is one of the biggest limitation of Kulum's law that it can be used to determine the force between two point charges or the electric field intensity of point charges which are static in nature.

[00:05:30]But what about the charges like these?

[00:05:36]If you have line charge charge ac spread across a lineed one-dimensional charge charge spread across a surface arbitrary surface surface means two dimension so just like this surface charge charge spelled across a volume three dimensionality spread volume charge now this kulum's law cannot be applied to determine the electric field intensity due to these types of charge distributions line surface and volume charge. Essentially kulum's law is valid only for point charges. So that's one of the biggest uh limitations. There are several other small things but mathematical difficulty everything else but the ba basic main limitation of the kulum's law is that it is applied to only point charges for charges like these line surface and volume kulum's law has a limitation let's see now What is Goss law of electrostatics and what it can do? What is the advantage? So when we discussed the limitation of Kulum's law that it is valid only for point charges the advantage the merit of the Goss law of electrostatics is that it is applicable to any type of charge distribution to determine the electric field of any type of or due to any type of charge distribution.

[00:07:24]So what it states is electric flux through a closed surface enclosing a charge is 1x epsilon KN times the charge enclosed you know epsilon KN is the permitivity of the free space epsilon times the magnitude of charge enclosed. So let's see break break break this statement down into two parts is equals to 1x epsilon * the charge enclosed 1x epsilon * the charge charge enclosed is q 1x epsilon * the charge means q by epsilon and what is that equal to the flux through a closed surface flux through a closed surface so this ph is equal to q by epsilon or not in in simple terms let's uh look at this let's say we have a charge Q okay and enclosing that we have enclosing that we have an arbitrary closed surface now whenever there is a charge you know to a positive charge the electric lines of force are always radiated outward means electric lines of force are the electric flux.

[00:08:50]Okay, how much of flux is coming out through a closed surface? That's what I want to know. Five, if you want to know how much flux, you should be knowing how much charge is enclosed. It means to say if you have more charge, if charge is more uh automatically the flux coming out if it is a positive charge because flux due to a positive charge radiator radiates outward. So if charge is more flux is more. So to know how much flux is radiating outward you should be knowing how much of charge that closed surface is enclosing. So essentially the go law of electrostatics states the flux through a closed surface enclosing the charge is equals to 1 by epsilon times the magnitude of the charge. Let's look at what this is. 5 is equals to integral over the closed surface. If we put a circle to the integral, it is a closed surface eds.

[00:10:10]Let's now look at the proof of that statement.

[00:10:15]One should be knowing that the electric flux D5 means elemental flux flux coming out to a small surface. If the surface is small dux is also small. So this one you are seeing here that's what is the elemental surface that we have cons considered.

[00:10:42]If the flux if you consider the flux coming out through that small surface the magnitude of flux also will be small. Area is small the magnitude of flux is also small. Area is large. If you increase the area the flux coming out is also large. It means to say the flux is proportional to area.

[00:11:10]And also the flux is also proportional to the electric field due to this charge.

[00:11:19]When you keep a charge definitely there will be electric field radiated due to it because you have a charge electric lines of force will be radiating outward. If it is a positive charge so if you have electric lines of force definitely you have electric field. So the flux d5 is equals to e ds.

[00:11:41]Now it is not the simple mathematical multiplication eds. It is e do ds just like work done is equal to f do s. So if you take away the dot you'll get cos theta. You know f do s or a dob is a b cos theta. uh just like that dy is equals to e d s cos theta. So what what we are trying to do is we are trying to get uh what is phi equal to we are first seeing what is d5 then we can integrate it then you'll get the total flux. So we are trying to see the flux radiating through an elemental surface ds. So that is d5 is equals to eds.

[00:12:28]Also we know that um E is equal to 1x 4 pi epsilon q by r². You know how we arrived at this formula. E is equals to f by q force through the kum law force kum's law has q1 q2 by r² charges will be there out of which one is a unit positive charge divided by q.

[00:12:54]So one of the charge gets cancelled. So what remains is is equals to 1x4 pi epsilon qxi r².

[00:13:03]You um work on that f is equal to q1 q2x 4 pi epsilon r² 1 q gets cancelled because of the denominator. So this value of e is substituted in this expression. Therefore it leads to d5 is equ= to qx 4 pi ep r² d s cos theta.

[00:13:29]Now just rearranging the terms what I'm doing is ex 4 pi epsilon kn at one place d s cos theta by r² this d s cos theta by r² has significance.

[00:13:43]So let's see what is uh that ds cos theta by r².

[00:13:51]Now if you see here the charge Q at the center of this closed surface uh the distance to the surface is R and uh what we have done is the electric field is radiating outward that is E and DS is a elemental uh surface uh I've drawn a normal to it that is DS.

[00:14:18]So ds and e need not be always parallel to each other. Only for a spherical surface both ds as well as e will be parallel to each other. This being an arbitrary surface, what happens is if I consider a surface like this here, uh this might be the electric field and uh this might be the DS vector normal drawn onto the elemental surface which we have considered. So only for a spherical surface let's say if you have a uh symmetric distribution your electric field will be this and if you consider an elemental surface and a uh normal to it both will be parall to each other otherwise there will be an angle made now so uh you know what is theta now what is e what is ds it's the normal projected to that elemental surface q is the charge charge and r is the distance between the charge and the surface that we have considered.

[00:15:27]Now if you uh see the angle subended due to this um uh elemental surface which is white in color if you observe that is d omega.

[00:15:41]Okay that's the solid angle subended uh due to the elemental surface which we have considered. Now let us say I have increased the elemental surface.

[00:15:55]Therefore the cone uh that is projected from the charge magnitude also increases means the d omega also increases. Now if I consider much bigger area and what happens is the cone d omega angle subended at the cone that's the solid angle also increases.

[00:16:27]So that value d omega is equal to d s cos theta* r² which we have isolated. There is a rigorous uh mathematical procedure to find out the solid angle subended uh by the surface at Q. It's it's essentially the cone cone starts widening up as you consider more and more area. What happens is if you integrate D omega uh you are trying to initially the solid angle subended small then increases increases. If you consider the total surface area of the closed surface then it becomes a total solid angle. Okay. Initially what we are doing is in place of ds cos theta by r² you have written d omega and the rest 2x4 pi epsilon kn remains the same. This is from previous expression. Initially itself I told you that we are trying to develop the expression for phi uh which is a flux coming out um through the closed surface. So we integrate phi is equals to integral d5 and that is integral uh this value integral of um by 4 pi epsilon d omega. You know the value of d omega and what is d omega? D omega is the uh solid angle subendered by the surface at Q the surface increases if means DS increases continues to increase so that it covers the entire solid entire surface closed surface then it it's called as a total solid angle.

[00:18:10]So what happens is this qx4 pi epsilon kn is constant comes out right integral d omega integral d omega is 4 pi means the total solid angle subended by the entire surface arbitrary surface uh is equals to 4 pi and again here there is um mathematical proof uh for that integral d omega what is the total solid angle subended that value is equals to 4 pi we are taking that value directly in such case the 4 pi 4 pi gets cancel uh what remains is the statement which we have seen initially that is phi is equals to q by epsilon the flux coming out through the closed surface is 1x epsilon kn times the charge enclosed uh in the simplest terms if you want to remember you want to know how much flux is coming out, you should be knowing how much of charge is enclosed. So simple it is means both are proportional to each other.

[00:19:23]Now we want to write uh it in the integral form.

[00:19:28]What I'm doing is this expression you know uh is goth law of electrostatic. I want to do it in the integral form. integral means the left hand side is written in integral form where it is written with integration the right hand side is not so I want to convert the right hand side also in the uh integral form what I'm doing is if this epsilon kn is taken to the other side something like cross multiply it goes to the other side like this epsilon kn into e is d so the same thing can be written like this. If you want to write in terms of D uh it's electric flux density or we also call it as a displacement vector in in dialectrics.

[00:20:21]So our aim is to um convert that into integral form is the the right hand side also should be written in with the help of integration. So we all know that q is equal to integral v row v dv. It comes from this density is mass by volume.

[00:20:39]Charge density is charge density I write it as row V. Charge density is charge by volume. Uh charge is uh Q by volume is V.

[00:20:58]Okay.

[00:20:59]If uh there is elemental volume elemental charge if you cross multiply dq will be equals to row v dv.

[00:21:13]If you want the total charge integrate if you integrate this is what we get.

[00:21:22]So uh volume charge density is dq by dv.

[00:21:25]So if you multiply row v dv dq is if you want to integrate that integral dq is q and integral row v dv is what is remaining. So this value of I want to substitute here so that the right hand side also is written in the form of uh with the help of integration. So it becomes the integral form of god's law.

[00:21:51]Now if you observe uh that's what is written integral e ds um if you cross multiply to the other side epsilon not e becomes d if you don't cross multiply this is what the main equation we have so integral row v by epsilon actually it is q by epsilon kn but q we have written as integral row v dv okay and if If you cross multiply to the other side we can write in the form of d integral dds is equals to provid dv. So either this equation in terms of e or equation 8 in terms of d both represent the integral form of gas law. Essentially the Goss law is um flux is equals to integral uh EDS for sure uh that is equals to Q by epsilon KN instead of this uh Q we have written integral row V DB and U that becomes integral row v by epsilon kn into uh dv. If epsilon kn is taken to the other side it becomes d. So what we have seen now is uh we have proved uh the mathematical form of go law and then we have converted that gla mathematical form of gola of electrostatics into the integral form. Essentially integral form uh sees that the left hand side and the right hand side is both represented in the with the help of integration essentially used for solving um complex uh scenarios where charge distributions are different.

[00:23:57]Now we uh try to bring it to the differential form where uh um both left hand side and right hand side is uh represented in the form of uh differential uh equation like d by dx.

[00:24:12]So um this is what the integral form we have we have seen just now.

[00:24:19]Now we are using this gau divergence theorem. Essentially the God's divergence theorem gives the relationship between surface and volume integral. Let a be any vector we can write as integral over the closed surface a dx a is any vector is equals to over the volume dell do a db d is um the differential operator uh that is equals to i do by dox plus j do by dy plus k do by doz.

[00:24:57]So um it is a differential operator where it can be operated on any vector if we have because you know d is i into d by dx. So it does that differentiation plus j do by dy plus k do by dz. So we call it as a differential operator dell.

[00:25:17]So we write it as dal dell.

[00:25:22]Now what happens is um if you see the left hand sides are same what we can do is equate the right hand side. So what happens in our next slide you'll see that uh these two will be equal del do e is equal to row v by epsilon kn.

[00:25:42]Now we observe here that del do e is equals to row v by epsilon kn. As usual, if you take this epsilon KN to the other side, which is a constant, we can write delts E is equals to row V.

[00:25:59]You know, epsilon E is uh D. Uh therefore we write as del E is equals to row V. So just like in case of the integral form we have written both in terms of e as well as both d in terms of e as well as d. So del dot e is equals to row v by epsilon del do is equals to row v. Now you see equation three and four we call it as differential form of G law of electrostatics or we also call uh it as a point form of uh gas law of electrostatics. So if you see here uh what exactly this dell dot dell dot represent it is the product dell dot or dell do e or dell dob both represent.

[00:26:58]So you know dell is a differential operator as I mentioned I do by dx plus j do by dy plus k do by doz and what is essentially e can be written as you know electric field is a vectoral quantity.

[00:27:18]So e is a vector. Uh we write as e ex of i.

[00:27:25]E is the x component. Um and this is a general form component may exist or not. Uh it all depends upon the scenario which type of charge distribution where it is uh ey of j and uh then it becomes uh ez of k ez of k. When we take uh a dell uh when it is operated upon the electric field uh it becomes I do by dx dot ex I so I dot i is 1 j dot j k what happens is do by dx of ex plus d by dy d of dy plus d by doz of ez. So what we are doing is uh through this dell we are doing the partial differentiation we are trying to see how the electric field varies with respect to x-axis or along x-axis along yaxis and along z ais that that's the meaning of uh the dell uh the partial differentiation do by dx do by dy and do by dozuh previously that um kulum's law is limit limited to only point charges. Whereas we have this Goss law which can be extended to uh apart from point charge line surface and volume charges also.

[00:29:14]The advantage lies in uh we construct a Gaussian surface and um we have uh we consider a elemental surface over that uh Gaussian surface. That Gaussian surface can be a spherical Gaussian surface, a cylindrical or a cube cuboid type of surface on to which we consider an elemental surface and then we integrate um once we integrate we'll get the total electric field uh due to that type of charge distribution be it uh line surface or volume type of charge distribution. So that's the advantage of the Gaussian uh Goss law of electrostatics. The only thing is one should be able to properly identify the Gaussian surface and uh once Gaussian surface is identified an elemental surface over which uh the ds is projected as well as e. So what is the uh direction what is the angle between electric field and ds is needs to be estimated.

[00:30:22]This is what the lecture summary what we have seen till now. Uh we have seen the proof uh of Goss law of electrostatics.

[00:30:31]Uh how ph is equals to q byepsilon not mathematically we have seen uh apart from the statement and also we have converted that mathematical form of go law of electrostatics into the integral and differential form. So both this uh integral and uh differential form are very much useful in solving uh problems.

[00:30:54]Integral form essentially uh where both the left hand side and right hand side um in integrals and differential form where we we see the partial differentiation of how the component of electric field varies along xy and z axis respectively.

[00:31:17]So in our uh next video I'll try to explain uh how to use this goss law of electrostatics uh for at least uh three different types of charge distribution. we start with point charge um and we'll check whether it leads to uh the the same electric field intensity that we obtained by Kulum's law and then we extend it to the uh line charge as well as the surface charge