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### 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

### 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 magnitude0128

which implies then that the total flux is going to be the integral about little bit of electric flux0147

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 da0309

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 cases0554

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

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 and0656

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 surface0705

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/ E₀0884

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

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 things1079

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 total1183

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 bigger1308

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 and1436

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 and1601

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 lesson1827

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 seconds1863

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 + Q1886

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 there1907

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 R²2321

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

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 tricky2542

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 and2565

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 law2706

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 shell2718

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 surface2744

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