For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Related Articles:

### Gauss's Law

- Electric flux is the amount of electric field penetrating a surface.
- Normals to closed surfaces point from the inside to the outside.
- The total flux through a closed surface is positive is there is more flux from inside to outside than outside to inside.
- Gauss’s Law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the material enclosed.
- Gauss’s Law is useful for finding the electric field due to charge distributions for cases of spherical, cylindrical, and planar symmetry. Gauss’s Law is always true, but typically only useful in these symmetric situations.

### Gauss's Law

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:07
- Electric Flux 1:16
- Amount of Electric Field Penetrating a Surface
- Symbol
- Point Charge Inside a Hollow Sphere 4:31
- Place a Point Charge Inside a Hollow Sphere of Radius R
- Determine the Flux Through the Sphere
- Gauss's Law
- Total Flux
- Gauss's Law 9:10
- Example 1 9:53
- Example 2 17:28
- Example 3 22:37
- Example 4 25:40
- Example 5 30:49
- Example 6 45:06

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### Transcription: Gauss's Law

*Hello, everyone, and welcome back to www.educator.com.*0000

*I'm Dan Fullerton and in this lesson we are going to talk about Gauss’s law.*0003

*In our last lesson, we spent quite a bit of time doing derivations of electric fields due to continuous point charge distributions.*0008

*Gauss’s law is one of my favorite lessons in Physics.*0016

*After we have gone through that really tough lesson and you survived it, we are going to show you a much simpler way to deal with a bunch of those situations.*0019

*Our objectives, number 1, understand the relationship between electric field and electric flux.*0028

*We will start by talking about what electric flux is.*0033

*Secondly, calculate the flux of an electric field to a variety of surfaces.*0037

*State and apply the relationship between flux and lines of force.*0041

*State Gauss’s law in integral form and apply it qualitatively.*0045

*Apply Gauss’s law to determine the electric field for planar, spherical, and cylindrical asymmetric charge distributions,*0049

*that is going to be a big portion of the lesson.*0056

*Almost without a doubt that is going to be involved in at least one of the free response questions on your AP exam.*0058

*Finally, apply Gauss’s law and determine the charge density or total charge on the surface in terms of the electric field near that surface.*0066

*Let us start off by talking about what is electric flux.*0074

*Electric flux is the amount of electric field penetrating a surface, we will get it the symbol Φ.*0079

*If we have some random surface, whatever surface happens to be,*0086

*what we will do is we will define a small piece of that surface, and from that, coming out of it perpendicular to it, we will define a vector da.*0090

*Where the direction of da, the area is perpendicular to that surface.*0105

*Assume we have some amount of electric field coming out of that surface as well, coming out through that area.*0108

*The angle between these we will call Θ.*0117

*Then this little bit of electric flux coming through that red portion, coming through our little bit of da,*0122

*we will define as dΦ the flux = the electric field dotted with that area vector da or E da cos Θ for dealing with the magnitude*0128

*which implies then that the total flux is going to be the integral about little bit of electric flux*0147

*which is going to be the integral / the entire area of all those little bits of E da.*0156

*Almost think of this like airflow or rain on a cookie sheet.*0165

*You have a cookie sheet held flat and it is raining on it, the amount of rain that is hitting that cookie sheet would be equivalent to the flux.*0169

*Flux goes through the surface that does not work with rain and cookie sheet, it made up the idea.*0177

*If you start tilting it in different directions, different amounts of rain hit the cookie sheet or go through it, went through a great perhaps, something like that.*0181

*What happens if you have a close surface instead of just this open surface?*0193

*Let us draw a close surface, something like that.*0198

*You will have to try and picture that in 3D because my drawing skills are poor at best.*0203

*In this case, we could define some little piece over here as da.*0211

*You could define say something over here as da.*0220

*Once again, the total flux is going to be the integral over the close surface, a shorthand notation integral over a closed surface of E⋅ da.*0229

*You are integrating over the entire thing to enclose it.*0243

*By convention, normal to close surfaces point from the inside to the outside.*0246

*Total flux through a close surface is positive.*0254

*If there is more flux from inside to outside, it is positive if there is more going out than going in.*0257

*If more is going than coming out, we would call that a negative flux.*0262

*This is just a convention as far as science go.*0266

*Let us take a look now at a point charge inside a hollow sphere to help us get an idea of how the Gauss’s law thing really works.*0270

*We will place a point charge inside a hollow sphere of radius R, let us draw ourselves a nice happy little sphere of some radius R, in its center we are going to put a point charge.*0279

*Let us make it +q.*0295

*It is pretty easy to see that electric field lines are going to be symmetric due out, assuming I can draw a little bit better sphere.*0298

*We could then say that the little bit of electric flux through some little piece of this da*0309

*then we define that da here is going to be equal to the electric field dotted with da.*0319

*which as we said previously is just E da cos Θ.*0331

*If it is a point charge inside a hollow sphere, anywhere in the electric field goes through that sphere,*0338

*it is going to be perpendicular uniformly around the entire sphere.*0343

*That means that Θ, everywhere the angle between those two was going to be 0.*0347

*cos of Θ is going to be = 1.*0354

*Therefore, the differential of that total flux, that little bit of flux is going to be just E da.*0358

*Our total flux then it is going to be the integral of all those little bits of flux or the integral over this close surface around the entire sphere of E da.*0367

*Since it is a point charge inside the center of a sphere the electric field is going to be the same everywhere as you go around the entire area.*0381

*For the purposes of this problem, E is a constant so that E comes out of the integral sign, E integral over the close surface of da.*0389

*The integral over the close surface of that differential of area, as you go around the entire surface,*0399

*it is just going to give you that entire area so that becomes Ea.*0404

*And as you know the area of a sphere = 4π R².*0410

*We can write then that the total flux is going to be = E × 4π R².*0421

*Even further though, we know the electric field.*0435

*We just spent a whole lesson on it.*0439

*The electric field due to that point charge = Q/ 4 π E₀ R² in the direction of r ̂ coming out from that point charge.*0442

*We could then write the flux as being = we have got our 4π R² × our electric field Q/ 4π E₀ R² and all of that then must be =,*0458

*Let us see, 4π 4π R² R² = Q/ E₀, simplifies very nicely.*0486

*Which implies then that the total flux = the integral/ the close surface of E⋅ da which was equal to the total charge enclosed by our sphere ÷ E₀.*0499

*There is a poor mans derivation of what we are going to call Gauss’s law,*0518

*the first of Maxwell's four equations which form the backbone of this entire E and M course.*0525

*Gauss’s law, you got to know it front words, backwards, upside down, and underwater.*0531

*The total flux is the integral over the close surface of E ⋅ da which is the total charge enclosed by your surface ÷ E₀.*0536

*Looking at this in a little bit more detail, what is this law good for?*0549

*It is extremely useful for finding the electric field due to charge distributions in cases*0554

*where you have symmetry and you need specifically 1 of 3 types of symmetry.*0559

*You need spherical symmetry, planar symmetry, or cylindrical symmetry, for this to be a very useful law.*0563

*Gauss’s law always holds, it is always true.*0570

*However, it is really only useful for the most part in these cases.*0573

*What it states, the total flux is the integral/ the close surface of E ⋅ da which is the total charge enclosed by that surface ÷ Ε 0, that constant.*0578

*Let us do some examples.*0591

*Find the electric field inside and outside a thin hollow shell of uniformly distributed charged Q.*0594

*Let us see if we can draw this first.*0602

*Find the electric field inside and outside a thin hollow shell of uniformly distributed charged Q.*0605

*There is our shell of charge with some amount Q.*0615

*What we are going to do is we are going to choose a Gaussian surface and vision this imaginary close surface,*0621

*in this case it is going to be a sphere to give us the most possible symmetry.*0630

*First we are going to do that inside the shell then we are going to do it outside the shell,*0633

*whenever we want to know the electric field strength.*0638

*By symmetry, the electric field at all points on this Gaussian surface, on this Gaussian sphere must be the same and must point radially in or out.*0640

*As we do this, let us start with inside.*0650

*Let us look inside this sphere.*0654

*We will define a Gaussian sphere here in green, this mental sphere that we have made up to help us solve the problem and*0656

*we will say that at some radius that we will call Ri which is inside the radius of our shell of charge there in blue with radius R.*0663

*Here is our Gaussian surface in green and we are looking for the electric field inside.*0677

*Where Ri is less than R.*0684

*Let us even spell that out Ri is less than R for this analysis.*0686

*Once we do that, we can say that the integral over the close surface of E ⋅ da = Q enclose by Gaussian surface ÷ E₀.*0693

*Gauss’s law we start off by writing it, then we also know that the area over here of our Gaussian surface*0705

*is just going to be 4π Ri² because we picked a sphere inside the shell of charge.*0717

*We have as much symmetry as possible.*0726

*Then we can state that the left hand side E is constant everywhere.*0729

*The integral of da is just going to be the area, left hand side becomes ea = Q enclosed/ E₀.*0735

*We did not have to do any real integration when you are using Gauss’s law, vast majority of the time if you have to do any real integration over here,*0743

*you probably did not pick the greatest surface.*0750

*Ea = Q enclosed/ E₀ which implies then that E × our area 4π Ri² = the charge enclosed ÷ E₀.*0753

*Or we can say then that E = charge enclosed / 4π E₀ Ri².*0769

*What makes this one especially easy though is the charge enclose by our Gaussian sphere is 0,*0784

*all the charges outside in this blue shell of charge there is nothing inside the Gaussian sphere.*0789

*Q enclosed = 0 therefore, the electric field = 0 inside that hollow shell of charge.*0794

*Gauss’s law defined the electric field inside where the 0 everywhere.*0807

*Let us take a look at the second area.*0812

*We are going to look when we are looking outside that hollow shell of charge.*0814

*If you are looking at the green one, let us take a look at this purple one.*0819

*We will set a Gaussian surface, another sphere, that hollow shell I should say, outside our shell of charge.*0823

*We will call this one Ro, as radius Ro.*0834

*We are looking where Ro is greater than R or outside that hollow shell of charge.*0841

*We will start off by rating Gauss’s law again, integral / the close surface of E ⋅ da = charge enclosed ÷ E₀.*0850

*The same thing again and the same derivation as well, E is a constant everywhere because we chose our Gaussian surface very carefully.*0864

*The integral over the close surface of da is just going to be a, so the Ea = Q enclosed/ E₀.*0873

*Where this area is the area of your Gaussian sphere, the area of that sphere is going to be 4π Ro² = Q enclosed/ E₀*0884

*which implies then that the electric field is going to be our Q enclosed/ 4π E₀ R 0².*0906

*The total charge enclosed + Q that is going to be Q ÷ 4π E₀ R 0².*0922

*For any R that is an Ro, that is outside this hollow shell of charge.*0934

*This is the same as if all the charge was placed in a point in the center of the sphere.*0942

*You could treat it like it is a point charge once you are outside, it does not matter.*0947

*What shape it is, as long as you can use Gauss’s law in it, this could be any amount of charge, any size diameter shell,*0951

*as long as you are outside you can treat it all as if all of that charge was condensed in a single point inside the center of the shell.*0958

*See how powerful this is? how much slicker this is?*0967

*You can just see the beauty and all the symmetry of the math as you do these sorts of problems.*0972

*We will go through a couple more examples to help clarify it but let us make a graph quickly of the electric field vs. radius just to help finish this one off.*0976

*There is our radius R, there is our electric field, and what we have is inside that value of R, let us make this R,*0992

*I will write it in blue even so it corresponds to our shell of charge here.*1003

*There is the radius of our shell of charge, inside that the electric field is 0 and outside that is Q/ 4π E₀ R 0².*1009

*What we have is a discontinuity where we got something like this, where that is proportional to α sign or fish sign, whatever you want to call it.*1020

*It is proportional to 1/ R² as long as you are outside R.*1030

*There is your graph of electric field vs. radial distance.*1037

*You can use Gauss’s to help you find these so quickly and so easily.*1040

*Let us do another example.*1046

*Find the electric field due to an infinite plane of uniform charge density Σ.*1049

*We have got an infinite plane, let us draw that here.*1055

*We are applying a charge that goes on and on and on and we have given it sum surface charge density Σ which is total charge ÷ area.*1063

*Since it is infinite, we cannot know the total charge or areas.*1072

*We will find a ratio of those and define the Σ.*1075

*Choose our Gaussian surface, you got to be smart about this and having done these sorts of things*1079

*before I know that a good way to do this is to pick a cylinder.*1085

*I'm going to pick a cylinder, I think almost of a soda can, centered that the plane.*1089

*There is the top and it also extends to the same amount underneath.*1107

*There is our Gaussian surface, a cylinder and we will set it so that it is some distance D above and D below that infinite plane.*1112

*By symmetry, the electric field at all points on the cylinder must point perpendicular to the plane through the caps of the cylinder.*1127

*On the sides, it should be pretty easy to see by symmetry everything is going to cancel out,*1134

*We are only going to end up having to worry about these caps our cylinder but let us go through that step by step.*1138

*Let us start by writing Gauss’s law, the integral / the close surface of E ⋅ da = our enclosed charge ÷ E₀,*1146

*which implies then since we know Σ is Q/ a and Q = Σ a.*1161

*We can write then the left hand side, this total flux I'm going to break up into pieces.*1170

*I'm going to take a look at the flux through the top, the flux through the sides, and the flux through the bottom cap.*1176

*That is the flux through the top + the electric flux through the bottom + total electric flux through the sides to give me my total*1183

*which is this left hand side and that must be equal to our charge enclosed is Σ × a, where that is going to be a ÷ Ε 0.*1201

*Of course, by symmetry we can look and see that the flux to the sides of this are all going to cancel out so the total flux through the sides is going to be 0.*1220

*We can say then that the flux through the top + the electric flux through the bottom must = Σ a/ E₀.*1229

*The flux through the top and the flux to the bottom must be the same because they are symmetric.*1248

*This one was pointing down and this one was pointing up.*1256

*We are worried about magnitudes for the time being.*1257

*The flux through the top = the flux through the bottom which by the way has to be the electric field at that point × the area so that = Ea.*1260

*Flux through the top Ea + flux in the bottom Ea gives us 2 Ea = Σ a / E₀ or solving for the electric field then, the electric field just = Σ / 2 E₀.*1275

*There is no dependents whatsoever on D and we have seen this before when we talk about infinite planes and infinite lines, it does not matter.*1295

*Since, it is infinite the only thing you are worried about is that surface charge density.*1303

*How much simpler is that in the derivation that we did with that disk that we grew bigger and bigger*1308

*and bigger in the last lesson to figure out the electric field due to an infinite plane.*1313

*It is much slicker derivation right here using Gauss’s law.*1318

*Let us do a graph of the electric field vs. distance again just so that we are consistent.*1322

*There is D, there is our electric field, and since there is no dependence on D, our graphs looks like a nice straight line at Σ/ 2 E₀.*1335

*Very good, let us see if we can extend this a little further and talk about parallel plates.*1353

*Find the electric field outside between 2 opposite recharged parallel planes or plates.*1360

*Let us draw our plates first, we will draw a top plate, we will draw a bottom plate.*1365

*We will say that this one has charged + charge density + Σ.*1381

*This one must have - Σ as its charge density.*1385

*What we are going to do is we are going to look at the electric field due to the different plates in different regions.*1390

*If I look at the electric field due to just the top one, the + Σ over here on the left,*1396

*the electric field due to the top plate must be going away since that is a positive charge.*1402

*On this side of it must be going away and over here, we are still looking at the electric field just due to this one so it must be going away.*1408

*Everywhere the magnitude of that is Σ / 2 E₀.*1417

*Let us take a look at the electric field due to just the bottom plate in these regions.*1424

*As a negative charged density it is going to have the field lines going toward it.*1430

*Up in this region we have Σ/ 2 E₀ toward it, the same thing here still toward the bottom plate and*1436

*we are looking at just the electric field due to the bottom plate for now.*1446

*And down here we are going toward that negatively charged plate.*1449

*When we put these together, if we want to find the total or the net electric field up above the 2 plates,*1453

*we have Σ / 2 E₀ up and Σ/ 2 E₀ down which gives us an electric field = 0 there.*1460

*In between the 2 plates, we have Σ / 2 E₀ down due to the top plate.*1468

*Σ / 2 E₀ down due to the bottom plate so our total electric field here is going to be Σ/ E₀ headed down.*1473

*At the very bottom, we have Σ/ 2 E₀ down, Σ / 2 E₀ up, so the electric field is 0.*1488

*What we really find after all of this is, is that between the parallel plates above the plates no electric field,*1496

*below the plates no electric field and between the plates Σ/ E₀ going from the positive to negative.*1504

*It is important to note that this is not accurate when you get near the ends of real plates.*1512

*If it is an infinite plane of charge, absolutely works great.*1517

*When you get to the edges you actually get some sort of fringe in effects, like this.*1520

*You got to have very big plates and either neglect the edges or have infinite plates for this to make sense.*1525

*But it is a pretty good approximation for most plates especially in the region not near the very edges of those plates.*1530

*Let us take a look at another example.*1538

*Find the electric field strength in the distance R from an infinitely long uniformly charged wire of linear charge density Λ.*1542

*We have done wires before, if I recall we got quite a bit of math and be quite a bit simpler here using Gauss’s law.*1550

*Let us start by drawing our wire and we are going to give it some linear charged density Λ.*1558

*We have got to pick a Gaussian surface that is going to give us the most possible symmetry.*1573

*In this case for a wire, what I'm going to choose is a cylinder that is centered on that wire.*1580

*Let us choose a cylinder, something like that for our Gaussian surface.*1587

*What is nice there is we got a cap on this end, a cap on this end.*1598

*Just by observation we can see by symmetry that those are going to completely cancel out and*1601

*all we are going to have to worry about is the pieces going through the sides of our cylinder,*1607

*through the edges of our can not the top or the bottom.*1612

*We will start off by writing Gauss’s law, the integral over the close surface of E ⋅ Da = the total enclosed charge, the charge enclosed by our Gaussian surface ÷ Ε 0.*1615

*This is the total flux but we are going to break that flux up into a piece through the left cap, a piece through the right cap,*1632

*and our piece through the sides I’m just going to write that as the cylinder portion for our flux must = Q enclosed / E₀.*1641

*As we just mentioned, by symmetry and we can state that the flux through the left + the flux through the right must = 0.*1655

*All we have to worry about is the flux through the sides of our cylinder.*1671

*When I'm calling cylinder portion = the total enclosed charge through our cylinder ÷ E₀.*1675

*To find the flux due our cylinder, first thing we have to realize is because it has the wire in the middle,*1683

*the electric field is going to be the same anywhere on the edges of that cylinder, anywhere around it.*1689

*That nice piece of symmetry again as we chose our Gaussian surface very wisely.*1694

*We need to know the area of that, however.*1700

*If we have the edges of the can to find the area of it, we really need to do is think about cutting it and spreading it out to.*1703

*We got a cylinder, we got to spread it out somehow.*1710

*What would the dimensions of that B once you spread it out?*1713

*The way we are going to do that is we will say that the flux through the cylinder is going to be,*1718

*if the radius of the top of the cap is 2 π R and we cut it,*1726

*that is going to be one dimension of our rectangle we get when we spread out that canned peas.*1730

*That will be 2π R and the other dimension is going to be some L that we have not defined yet, the length of our can.*1736

*We have got that and we have to multiply that all of course by our electric field to get the total flux.*1747

*There is our Ea which is the flux through that.*1753

*We can then write that we have our 2 π RL × our electric field = our enclosed charge ÷ E₀.*1757

*We will go back to our linear charge density Λ which we know Λ is total charge ÷ length.*1771

*Then our Q enclosed, the amount of charge enclosed by our cylinder is just going to be the linear charge density × this length L.*1781

*Then we have on the left hand side 2 π RL × our electric field = Q enclosed on the right Λ L ÷ E₀.*1793

*Or solving for the electric field E = Λ/ 2 π E₀ R.*1808

*Same as the answer we reached using that point charge summation in the previous lesson*1827

*that is so much simpler, easier, slicker, quicker, and much more intuitive too.*1833

*Gauss’s law is a very powerful tool.*1840

*Let us take a look at some of old AP questions to hone our skills here and we will start off by looking at the 2008 E and M exam free response number 1.*1845

*And you can look that up on www.google.com or college web sites, it is available for you to download and printout.*1858

*Here is a link to it currently, I would highly recommend taking a minute, printing it out, and taking as couple seconds*1863

*and seeing if you can solve it on your own before you check the answers.*1869

*If you get stuck that is perfectly fine then come back to the video and play along with it.*1873

*If you got the whole thing, just check your answers as you go through and see what you got right and what might need more work.*1877

*Use this is a practice opportunity.*1882

*In this question, we are starting off with a metal sphere that have radius A containing a charge + Q*1886

*and it is surrounded by an uncharged concentric metallic shell and a radius B and outer radius C.*1894

*It wants us to first determine the induced charge on the inner surface of the metallic shell.*1900

*The way I will start to do that is looking at party A here, we have some solid positively charged + Q piece there*1907

*and outside of that we have another conducting shell.*1918

*+Q in the middle and out here we are ask to find what is the charge on the inner surface of this shell.*1931

*Over here at this radius that they call B, that is B.*1938

*They are defining that as A.*1943

*The way I would do that to begin with is I would first choose Gaussian surface that is inside this shell and apply Gauss’s law.*1949

*I'm going to make a shell here that is actually contained, runs through the middle of that conducting shell.*1958

*There is my Gaussian surface, that sphere there in purple.*1968

*The integral over the close surface of E ⋅ Da as the enclosed charged ÷ E 0.*1972

*To do this next step, you probably need to have seen this before either in a previous Physics class.*1985

*You have to know that the electric field inside a conductor is 0.*1991

*When you know that the electric field inside a conductor is 0 this becomes very simple.*1996

*We will get that to that in a later lesson.*2003

*If you seen at a previous algebra based physics course that would be a great start here.*2005

*Once you know that then you can say the left hand side, if you know the electric field is 0, the left hand side 0 down with something must be 0 = Q enclosed ÷ E₀.*2012

*Therefore, Q enclosed by our Gaussian surface must be 0.*2025

*Q enclosed includes + Q from our inner sphere + let us call it QB, whatever we have on the inner surface of this shell must be = 0.*2032

*If Q + QB = 0 that implies then to me that QB must be = -Q. Our answer is –Q.*2052

*If you have + Q here on the outside of this sphere, you must have a -Q charge on the inside there using Gauss’s law.*2063

*The trick is knowing where to put your Gaussian surface.*2076

*Let us take a look at the second part there, find the induced charge on the outer surface of the metallic shell.*2080

*To do that, I'm going to choose a Gaussian surface now that is outside that shell.*2087

*I’m going to use that red Gaussian surface for part 2.*2093

*Integral now over the close surface of E ⋅ da which is Q enclose/ E₀,*2100

*always starting by writing Gauss’s law must be = the total enclosed charge must be + Q.*2107

*We know that because it says this is neutral in the hole and that is your positive + Q.*2115

*Once we do that, we can realize that Q enclosed = we have got our QA which we called + Q over there.*2122

*We have our QB and we now have our QC which is what we will call the charge on the outside of the sphere.*2139

*This implies then that our + Q, our total enclosed charge Q enclosed + Q must be = we know QA is + Q.*2153

*We know QB is -Q + whatever QC is.*2167

*The only way that will work, if we have + Q total, + Q and -Q is going to be 0,*2174

*That implies then that QC must be = +Q.*2179

*The charge on the outside here must be + Q which only makes sense by the law of conservation of charge.*2186

*If we said we had -Q on this inner surface and the whole thing is neutral we must have + Q on that outer surface.*2192

*There is how we would do part A.*2200

*Let us take a look and slide over to doing part B now.*2203

*It asks us to determine expressions for the magnitude of the electric field as a function of radial distance for different regions.*2210

*For the first region, we will call region I, we have R is less than A.*2217

*We are inside that inner sphere.*2224

*In order to do these problems, I'm first going to define now a volume charged density where Ρ = Q/V = Q/.*2228

*I'm not sure I'm going to need them because we got conductors 4π A³.*2242

*I will define it but I'm not sure we are going to need it now that I think about it.*2247

*Let us take a look here for are less than A, Gauss’s law integral / the close surface E ⋅Da = the total charge enclosed ÷ E₀.*2250

*The left hand side this becomes E × the area, the right hand side is our charge enclosed which is 0.*2264

*Therefore, we can state that the electric field is 0 inside that metal sphere,*2274

*which of course we knew because they already said just a few moments ago that the electric field inside the conductor must be 0.*2280

*We proved it.*2286

*Let us move on to part 2.*2289

*We are looking in that region between the solid sphere and that hollow conducting shell, or that solid conducting shell I should say.*2293

*Here we have R between A and B in that diagram.*2305

*The integral / the close surface of E ⋅ Da, Q enclosed / E₀.*2312

*Choosing a Gaussian surface that is in that region which implies that Ea which is going to be E × 4π R²*2321

*where R is between A and B must equal our total charge enclosed + Q ÷ E₀.*2330

*Therefore, the electric field was just going to be Q/ 4π E₀ R² in that region.*2339

*Just like it was a point charge.*2349

*Moving on to part 3, now we have R between B and C or again we are looking inside that region, we will draw a picture of it to make it clear.*2352

*There is our solid sphere, here is our conducting sphere, now we want our R somewhere in this region there in red.*2367

*We can do that easily in a couple of ways.*2379

*We know already since this is +Q here and –Q on this surface, our total enclosed charge is 0.*2382

*Therefore, the electric field is going to come out to be 0 or once again we are inside the conductor at this point electric field must be 0, straightforward.*2388

*Part 4, now what we are going to do is look for R greater than C.*2400

*We are in some region outside the entire object, the entire setup.*2406

*In this case, we will start off with Gauss’s law again, integral/ the close surface of E ⋅ Da = D enclosed charge ÷ Ε 0.*2413

*The left hand side becomes the electric field × the area E × 4π R².*2425

*Our total charge enclosed is just +Q, therefore the electric field again is Q/ 4π E₀ R².*2432

*Again, we are treating the entire thing as long as you are outside of that, outside of all those pieces, as if the entire charge was concentrated in a single point.*2444

*The same answer.*2453

*Again, the power of Gauss’s law.*2455

*It wants us to do a graph of the electric field as a function of distance.*2461

*It should be pretty straightforward because we just figured out all of the different pieces analytically.*2466

*Let us draw our graph here.*2473

*There is our R, there is our electric field, and let us make a couple marks here for A, B, and C, just to help us to limit where things are in the graph.*2485

*We already know between 0 and A the electric field is 0 and between B and C, the electric field is 0.*2508

*Those regions where you are inside the conductor electric field is 0 inside the conductor.*2516

*When we are in between A and B, we are proportional to 1/ R² and when we are outside of C we are again proportional to 1/ R².*2521

*Fairly straightforward and simple graph, that is part C of the question.*2536

*And part D, if this is your first Physics course it could be a little bit tricky*2542

*if you have seen had algebra based Physics before, then you probably had this before and it is a fair question.*2549

*Either way, it is fair by the time you finish the entire course sequence.*2555

*Given an electron of mass M and it is caring a charge - E is released from rest at very large distance from the spheres.*2559

*Find the expression for the speed of the particle at a distance 10 R from the center of the sphere and*2565

*they really meant to say 10 C from the center of the spheres.*2570

*What is helpful to know here is that the electrical potential energy due to those point charges is 1/ 4π E₀ × the one charge × the other charge ÷ R.*2576

*That is the piece that you probably have to know already.*2591

*I would do this from a conservation of energy perspective.*2594

*It is a fairly easy way to solve this problem knowing that the kinetic energy + the potential energy is going to remain constant.*2597

*Because you are starting at very large distance from the spheres we are going to say that is our potential energy level from rest to largest.*2608

*Since the rest kinetic energy is 0, potential of 0.*2616

*Therefore, the kinetic energy is going to be ½ ME V² and that is going to be equal to the potential energy Q × the charge E/ 4π E₀ × are distance 10 C.*2620

*Now it is just an exercise in algebra to find the velocity.*2644

*V² then is going to be equal to, we got the 2 there, we are going to have Q × E/ 2 π E₀ ME × 10 C.*2648

*Therefore, V itself is going to be equal to the square root of the QE/ 2 × 10 is going to be 20 π E₀ ME √C².*2667

*D might be a little tricky if you have not got in the potential energy piece and electrostatics yet.*2688

*We will get there in a couple of lectures.*2693

*The rest of it is pretty straightforward application and great practice for Gauss’s law.*2696

*Let us do one more free response problem here.*2703

*This one from the 2011 test free response number 1, that was kind of an interesting take on Gauss’s law*2706

*and whether you really understand how this law works and also a bit of close reading.*2714

*In this problem, they start off by asking you to use Gauss’s law to prove that the electric field inside a spherical shell*2718

*with the uniform surface charge of Σ outside surface and no charge anywhere else inside is 0 and describe the Gaussian surface you would use.*2728

*We only have been doing that for a while now.*2738

*Let us take a look, here we will call this A.*2740

*If we use a spherical shell that is inside the charged shell as our Gaussian surface*2744

*Then we have some charge Σ here and our Gaussian surface inside it.*2769

*The integral/ the close surface, E ⋅ Da = Q enclosed / E₀.*2780

*The Q enclosed inside there is 0.*2790

*Since Q enclosed = 0 we can say that E × the area 4π R² = 0.*2793

*Therefore, E = 0 nice and quick proof done.*2802

*For part B, it says the charges are now redistributed so that the surface charge density here is no longer uniform.*2809

*Is the electric field still 0 everywhere inside the shell?*2818

*The answer of course is no.*2823

*When you do that without the symmetry of charges to cancel each other out, there can be an internal electric field.*2826

*Gauss’s law still holds but it is not the same everywhere.*2833

*Using Gauss’s law is not overly helpful to you in this point for determining electric field.*2836

*I would write something like no, without the symmetry of charges to cancel each other out you may have an internal electric field.*2841

*Something like that should work.*2879

*They change that and here is where they throw the curveball, something you might not have seen before which makes it a really great question.*2884

*They give you a small conducting sphere was charged + Q with a center that is in the corner of a cubicle surface.*2892

*Again, if you have not downloaded and look at the question, make sure you do this.*2898

*Otherwise, this is not going to make any sense.*2902

*For which phases of the surface of any of the electric flux through the phase = 0.*2904

*As you look at the diagram, the only place the electric flux is going to be 0 is when the field lines are running parallel to the surfaces.*2910

*And of the 6 surfaces on the cube, 3 surfaces are going to be parallel, are going to be touching that points so you have electric field lines parallel to them.*2919

*No flux through those surfaces and those ones would be ADEH, ABCD, and ABGH.*2928

*It asks you to explain your reasoning, something like field lines run parallel to the surfaces.*2941

*Therefore, no flux thorough these surfaces.*2963

*Moving on, let us take a look at D.*2983

*At which corners of the surface does the electric field have the least magnitude?*2986

*That could be a really tricky if you do not read the question very carefully.*2992

*At which corner do you have the least magnitude of electric field?*2997

*You have to realize at this point that they are talking about a small conducting sphere whose center is at the corner A.*3002

*Conducting sphere inside a conductor the electric field is 0 so point A which is inside the conductor has to have 0 electric field.*3011

*I would state that A is inside the conducting sphere and the electric field is 0 inside the conductor at equilibrium.*3022

*It is kind of a trick question there if you have not read it very carefully.*3051

*Let us take a look at part E, find the electric field strength that the positions you have indicated in part D.*3057

*We just did that and I think they are trying to throw you off again by giving you all these fundamental constants.*3065

*You do not need them, you already know inside the conductor E = 0.*3070

*E at point A = 0 done.*3075

*Finally, last part of the question and last part of this lesson.*3079

*Given that 1/8 sphere point A is inside the surface, find the electric flux through phase CDEF.*3084

*Gauss’s law integral / the close surface of E ⋅ Da = Q enclosed / E₀.*3092

*Our Q enclosed is Q ÷ 8 because only 1/8 of that sphere is inside our cubed.*3103

*Q enclose is Q/8 so that is Q/ 8 E₀.*3113

*By symmetry, we have three faces of the cube that have flux running through them.*3120

*They are all symmetric, they all have the same orientation, the same distance with respect to that point A.*3127

*Therefore, each one is going to get 1/3 total flux so that is going to be 1/3 of Q/8 E₀ where I would say the flux to each one is going to be Q/24 E₀.*3134

*Hopefully, that gets you a good start on Gauss’s law, an extremely important part of the course.*3156

*If you struggle with it, please take your time go back and do some practice problems.*3160

*You are really going to need this for success in the APC E and M exam.*3164

*Thank you so much for watching www.educator.com.*3169

*We will see you again real soon, make it a great day.*3172

1 answer

Last reply by: Professor Dan Fullerton

Fri Feb 16, 2018 5:53 PM

Post by Kevin Fleming on February 16 at 03:07:13 PM

Professor Fullerton,

For example 5, question a, part ii, why did you equate the Q(enclosed) by the gaussian surface surrounding the outer radius of the metallic sphere to be +Q if, in part i, you said that the Q(enclosed) for the gaussian surface situated between the inner and outer surface was eqaual to 0. Wouldn't the +Q charge be neutralized by the -Q charge of the inner surface? Thanks

2 answers

Last reply by: Sunanda Eluri

Tue Sep 13, 2016 9:56 AM

Post by Sunanda Eluri on September 13, 2016

Hello sir,

If the radius of the gaussian surface is doubled, will it affect the flux passing through the surface?

1 answer

Last reply by: Professor Dan Fullerton

Sat Sep 3, 2016 3:58 PM

Post by isaac maingi on September 3, 2016

when moving the 4pir^2 from the left side of the equation to the right side of the equation to solve for E, whey does the 4pir^2 go to the denominator instead of the numerator?

4 answers

Last reply by: Professor Dan Fullerton

Mon Feb 22, 2016 6:17 AM

Post by Jim Tang on February 20, 2016

Hm... Same thing

In Example 4, my intuition told me the caps were 0 since dA and E were at 90 degrees...I think that's what you meant by "cancelling out."

In this case, it's not cancelling out between objects...but of simply cancelling out?

2 answers

Last reply by: Jim Tang

Sat Feb 20, 2016 9:02 PM

Post by Jim Tang on February 20, 2016

For Example 2, I think you have the wrong explanation for why the sides cancel out? My intuition tells me since cos(90)=0, the E will all be 0 along those sides. I can't see anything "cancelling out."

1 answer

Last reply by: Professor Dan Fullerton

Thu Feb 18, 2016 10:32 AM

Post by Daniel Jansson on February 18, 2016

Mr. Fullerton, I'd love some examples of Gauss law in differential form as well!

/Danny (Engineering physics student)

1 answer

Last reply by: Professor Dan Fullerton

Sun Apr 19, 2015 3:02 PM

Post by Patrick Jin on April 19, 2015

Hi! Thank you for the lecture first of all.

I have questions about the example 2. So when we use the cylinder shaped gaussian field for the infinite field, all the electric flux to the side cancel out.

Would it be correct if I say that up and down flux cancels out too just like the flux on the side, but since we are dealing with an infinite plane ( which technically only has top and bottom part with no left and right sides ) we are only considering the magnitude of the up and down flux ?

1 answer

Last reply by: Miras Karazhigitov

Sun Apr 5, 2015 7:52 PM

Post by Miras Karazhigitov on April 5, 2015

Hello, can you explain why in 5th example (1. a) ii ) the total charge is equal to +Q?

1 answer

Last reply by: Professor Dan Fullerton

Fri Apr 3, 2015 8:48 AM

Post by Luvivia Chang on April 2, 2015

hello professor, Can you explain why the flux through top and bottom are equal in the example 1 while the sum of the fluxes through left and right caps are zero in example 3? what is the diffeence between the 2 situations?

1 answer

Last reply by: Professor Dan Fullerton

Tue Feb 3, 2015 6:08 AM

Post by Arjun Srivatsa on February 3, 2015

Why does the electric flux of the sides of a cyclinder cancel out by symmetry in Example 2?

1 answer

Last reply by: Professor Dan Fullerton

Fri Nov 14, 2014 6:06 AM

Post by QuangNguyen VoHuynh on November 14, 2014

Also, I would like to know the reason why the electric filed in the conductor is zero? Is the reason because the total flux is zero?

1 answer

Last reply by: Professor Dan Fullerton

Fri Nov 14, 2014 6:05 AM

Post by QuangNguyen VoHuynh on November 14, 2014

Can you explain me the reason why we must use the cylinder when finding the electric filed in infinite plane using Gauss's Law?