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Lecture Comments (9)

1 answer

Last reply by: Professor Dan Fullerton
Fri Apr 8, 2016 6:23 AM

Post by Ayberk Aydin on April 8 at 03:06:11 AM

Also why can't you take the negative derivative of the electric potential with respect to dr to calculate the electric field in part e of the 2010 FRQ?

1 answer

Last reply by: Professor Dan Fullerton
Fri Apr 8, 2016 6:16 AM

Post by Ayberk Aydin on April 8 at 02:04:48 AM

Why is the electric field between .1 and .2 meters negative in the 2012 FRQ?

1 answer

Last reply by: Professor Dan Fullerton
Tue Jan 19, 2016 6:56 AM

Post by Shehryar Khursheed on January 18 at 02:30:56 PM

Also, another question I had was regarding the 2010 FRQ about the quarter circle charge distribution. This question is about the limits you chose for the integrals on both part b and part e. Firstly, I used the limits 0 and pi/2 for both because I assumed orientation didn't matter; only the length did. And for part b, I actually got the same answer as you even though I used different limits. Was this a coincidence or is it fine to use different limits like that when calculating electric potential?

Now, when I proceeded to do part e, I used the same limits aforementioned. However, now I got a different answer than you. I concluded that this didn't happen in part b because electric potential is a scalar. Since electric field is a vector, orientation does matter and my limits need to match the diagram. Is this correct? Thanks!

2 answers

Last reply by: Professor Dan Fullerton
Mon Jan 18, 2016 1:31 PM

Post by Shehryar Khursheed on January 18 at 01:07:07 PM

On the example calculating the electric potential of the disk, what happend to the little "r" in the numerator at around 7:45? You said we'd be left with an ri in the denominator but there was no mention of the r in the numerator of the integral. Thanks!

Electric Potential Due to Continuous Charge Distributions

  • E=-dV/dr
  • V=-Integral E dot dr

Electric Potential Due to Continuous Charge Distributions

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 0:10
  • Potential Due to a Charged Ring 0:27
  • Potential Due to a Uniformly Charged Desk 3:38
  • Potential Due to a Spherical Shell of Charge 11:21
  • Potential Due to a Uniform Solid Sphere 14:50
  • Example 1 23:08
  • Example 2 30:43
  • Example 3 41:58
  • Example 4 51:41

Transcription: Electric Potential Due to Continuous Charge Distributions

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

I'm Dan Fullerton and in this lesson we are going to talk about electric potentials due to continuous charge distributions.0004

Our objectives include calculating the electric potential on the axis of a uniformly charged disk.0010

Deriving expressions for electric potential as a function of position for some uniformly charged wires,0016

some parallel charge plates, coax cylinders and some concentric spheres.0021

Let us dive right in and let us start looking at the potential due to a charged ring.0026

Find the electric potential on the axis of a uniformly charged ring of radius R 0031

and total charge Q at point P located at distance Z from the center of the ring.0035

First, let us draw a picture of the situation and we have done this before for electric field.0041

That is going to look kind of like this.0047

There are our axis, we will call these Z, Y, and that would be X.0049

We will put our ring on it like that and we are going to look for the potential at some point P over there.0056

The first thing I’m going to do again is break this up into some little portions where we have got some Δ Q over here.0068

We will find the distance from Δ Q to that point P, there we go.0075

We will call that RI.0083

Now that we have done that, let us also mention that this radius is R.0087

To find the potential of point P, that is going to be the sum over all these little bitty pieces of their potentials 0094

which is 1/ 4 π ε₀ × the sum over all I of the charge contained in that little bitty piece QI ÷ RI,0105

which implies then the potential at point P is going to be 1/ 4 π ε₀.0119

As we make all of these little bitty pieces smaller and smaller and add them up, we can integrate.0124

Charge is going to be that little bit of charge DQ ÷ R.0130

But in this case, our R, as we go around the circle is going to be the same.0135

It is constant everywhere, RI = R.0139

Since we know that RI = R, we can say that the potential at point P = 1/ 4 π ε₀.0142

This is a constant, it can come out and we can replace it with RI integral of DQ.0155

Our integration just got pretty easy.0165

If we go all the way around the circle, adding up all the little bits of DQ, the integral of Dq is just our total charge Q.0167

We also can tell RI pretty easily using the Pythagorean Theorem.0181

That RI is going to be the √ Z² + R².0186

Therefore, our potential at point P is going to be Q/ 4 π ε₀ × 1/ √ Z² + R², our RI.0193

Pretty straightforward derivation for the potential due to that charged ring.0212

Taking a look at the uniformly charged disk we are going to follow the same strategy we did before.0218

Where we started off with a little diagram, we had our ring before and we are just going to expand that ring until we get a nice big disk.0223

We going to start our little ring small, make it bigger and bigger until we fill out the entire disk.0252

Where again, the charge of the little ring is going to be our DQ, R will be the radius of our entire disk, r will be the radius of our expanding ring.0256

As we do this, we will have some point over here P, as we draw from our little bit of DQ to P.0275

Let us do that in red so it stands out a little bit, there we go.0286

Our RI will be the √ C² + R² and that looks pretty similar.0292

The width of this is going to be DR, just like we did when you are doing the electric fields in a similar fashion.0302

Let us start off by taking a look at DQ is going to be 2 π R, the length of that hoop, 0308

imagine we cut it and pull it out, × its thickness DR, × our surface charge density σ, where σ is going to be charge ÷ area.0316

In this case that is going to be Q/ π R².0329

Then, we can do our substitution to say that DQ = 2 π DR × σ which is Q ÷ π R².0337

In just a little bit of simplification here, I got a π and π there so we can call this 2Q R DR ÷ R².0353

To find our potential, the electric potential at point P is the sum of all these little rings I and the potential of them.0367

The sum of all the potentials of those little rings I which is 1/ 4 π ε₀, sum of all of our dq/ RI, 0378

which implies then that the potential of point P is 1/ 4 π ε₀.0393

Making those tinier and tinier, and adding them all up using the integration process going from R = 0 0399

to R of our DQ which we said was 2Q R DR/ RI.0411

It is still down there and ÷ R², as part of our DQ.0420

We now have something that we can start to work with.0425

Pull other constants again, VP =, 2Q can come out and that R² in the denominator can come out.0429

I end up with 2Q ÷ 4 π ε₀ R², integral from 0 to R of, we are going to be left with our RI DR.0434

RI is √ Z² c+ R² in the denominator that is going to be Z² + R²⁻¹/2 DR, which implies then that our potential at point P will be Q/2.0458

Going from 2/ 4 to ½ π ε₀ R².0480

In order to do this, if this is our U, we need a D and we needed 2 over here, so we need a 1/2 there.0489

We will end up with ½ here, integral from 0 to R of Z² + R²⁻¹/2 × 2 R DR.0499

That is a form we can integrate.0519

Our R are variable so we get U ^½.0521

Our DU is over here.0524

Going to the next step, let us give ourselves a little bit more room here.0527

Potential at point P = Q/ 4 π ε₀ R² again, incorporating that ½ we had, integral from 0 to R of Z² + R²⁻¹/2 × 2 R DR.0533

Just making this a little bit clear, if we say that U is Z² + R² then DU must be 2R DR.0559

If we do that then that means that the integral of U⁻¹/2 DU must be 2U ^½ + Z.0571

Applying that to our integral VP = 2/ 4 π ε₀ R² × our integral here, 0585

we are going to end up with 2 × our U Z² + R² ^½ evaluated from 0 to R.0598

Because we have a definite integral, we do not need to use the +C.0614

Which is Q/ 4 π ε₀ R² × 2 × Z² + R² ^½ -2 × Z² ^½, or VP = Q, let us factor out that 2.0618

2 π ε₀ R² ×, that leave us √ Z² + R² – Z.0652

The electric potential at point P due to that uniformly charged disk.0672

Let us try the potential due to a spherical shell of charge.0681

Find the electric potential both inside and outside a uniformly charged shell of radius R and total charge Q.0685

Let us start with outside, the potential outside is going to be the opposite of the integral of E ⋅ DL.0694

Choose the opposite of the integral from infinity to R and we by now know the electric field equation pretty well, 1/ 4 π ε₀ Q/ R² DR, 0703

which implies that the potential outside = -Q/ 4 π ε₀ integral from infinity to R of R⁻² DR 0719

is going to be -Q/ 4 π ε₀ integral of R⁻² is going to be -1/ R evaluated from infinity to R.0736

That the potential outside = -Q/ 4 π ε₀, substituting in we have -1/ R - -1/ infinity that is going to be 0.0750

We get Q/ 4 π ε₀ R outside the shell.0765

If we want to look inside, it is going to be the opposite of the integral of the E ⋅ DL.0778

Again, it is the opposite of the integral from infinity to R of 1/ 4 π ε₀ Q/ R² DR, 0787

- the integral from R all the way to r, wherever we happen to be of 0 DR.0801

We are going to do that piece which means anywhere inside, that we do not have to worry about this we just did.0812

That is going to be Q/ 4 π ε₀ R.0818

Outside it is a function of how far where you are, inside it is going to be a constant.0826

If we want to graph that and I really think we do.0832

Let us take a look in plot potential versus distance from the center.0837

Let us put a little mark here for R the radius of our shell of charge.0852

This will be our potential, here we are.0860

Inside we have a constant which is Q/ 4 π ε₀ R.0865

Outside, however, we have Q/ 4 π ε₀ r.0875

It is proportional to 1/ R.0881

There is our potential due to a spherical shell of charge.0884

How about if we try a uniform solid sphere?0888

Find the electric field and electric potential inside a uniformly charged solid insulating sphere of radius R and total charge Q.0893

First thing that I'm going to want to do is to find the electric field 0902

and we will choose a sphere as our Gaussian surface and I'm going to put that Gaussian sphere inside that sphere of charge.0906

Use Gauss’s law to find the electric field first.0918

Integral/ the close surface of E ⋅ DA = our total enclosed charge ÷ ε₀.0920

The left hand side becomes E × the area of our sphere 4 π R² 0930

and that is equal to the charge enclosed, the volume charge density ρ × the volume ÷ ε₀.0935

As we do this, we have done some of the stuff before so we know that ρ is 3Q/ 4 π R³.0944

We have solved that previously.0961

The volume of the sphere is 4/ 3 π R³.0963

The left hand side becomes E 4 π R² equal to, we have got our ρ 3Q/ 4 π R³ × our volume V 4/3 π R³.0972

We still have our ε₀ in the denominator.0994

We got a couple of simplifications I think we can make here.0999

We have got a 3 here, we have got a 3, we got a π, we got a π, and I think that will do it.1002

We got a 4 and a 4.1010

That is equal to, we have still got a Q, we have got a R³ ÷, R³ in the denominator, 1013

we have got an ε₀ in the denominator, and we are dividing by the 4 π R² from the left hand side.1034

This implies then that our electric field = Q R/ 4 π ε₀ R³.1046

As we do this, it occurs to me that you know it is been a little bit since we did this.1065

Maybe we ought to talk about it for just a second about where that ρ = 3Q/ 4 π R³ came from.1068

If you want to take just a second and pull that out, for ρ = Q/ V, then that means that is charged / 4/ 3 π R³.1074

And that means that ρ = 3Q/ 4 π R³, since we know the total charge there.1088

If you are wondering where that came from, then we have done that before.1097

It is probably worth taking a second and doing that again.1102

We have got our electric field, next we are going to integrate to find the electric potential, 1105

noting that the total integration from infinity to R, our lower r has to be done piece wise 1110

because the electric field is discontinuous.1117

We are going to integrate from infinity to R and then from R to r because they are different functions.1120

Let us take a look at how we might do that.1129

Our electric potential = the opposite of the integral from infinity to R of E ⋅ DR.1132

Since we are doing this piece wise, V is equal to the opposite of the integral from infinity to R of E ⋅ DR –1140

the integral from R to r of E ⋅ DR in that region, which implies then 1152

that the electric potential is the integral opposite of the integral from infinity to R of our function.1161

Our electric field there which is 1/ 4 π ε₀ Q/ R² DR -, inside our sphere the integral from R to r of 1/ 4 π ε₀ R³ × QR DR, 1167

which is equal to, the left hand side is Q/ 4 π ε₀ R -, in the right hand side we can pull out some of our constants 1192

and come up with Q/ 4 π ε₀ R³, it can all come out, integral from R to R of R DR.1205

This is going to be equal to, we still got our Q/ 4 π ε₀ R - Q/ 4 π ε₀ R³ integral of R is R²/ 2 and that is evaluated from R to R.1222

All this becomes Q/ 4 π ε₀ R - Q/ 4 π ε₀ R³ × R²/ 2 - R²/ 2, which implies then, 1247

let us see if we can put some of this together.1273

V = Q/ 4 π ε₀ R -, let us do this separately -Q R²/ 8 π ε₀ R³, let us multiplying these through, 1275

+ R²/, that will be 8 π ε₀ R³.1302

All of that will equal, let us see what do we have here.1316

That is 2Q/ 8 π ε₀ R.1319

We will end up with 3Q because this becomes R²/ R³ we will gets 3Q/ 8 π ε₀ R combining this first and third term, 1324

-Qr²/ 8 π ε₀ R³, which implies then that the potential is equal to Q/ 8 π ε₀ R × the quantity, we will bring all that out, 3 - r²/ R².1347

The electric potential due to a uniform solid sphere.1376

Let us do a couple of AP free response problems and then we will move on here.1381

Starting off with the 2012 exam free response question number 1.1389

I have mentioned before, probably worth taking a few moment, downloading the question, printing I out, trying it, then come back here.1395

Pause buttons are wonderful thing.1404

We got two thin concentric conducting spherical shells insulated from each other a 3 DI of .1 and .2 m and it gives you a diagram of that.1409

Part A, using Gauss’s law, derive an algebraic expression for the electric field for R between 0.1 and 0.2 m, in that region, between those two.1418

For part A, we are going to use Gauss’s law.1429

The integral over the close surface of E ⋅ DA which is equal to our enclosed charge ÷ ε 0, 1433

which implies then that the electric field × 4 π R² = QI, that inner Q ÷ ε₀.1446

Therefore, the electric field is QI/ 4 π ε₀ R².1461

Moving on to part B, determine an algebraic expression for the electric field when you are greater than 0.2 m, 1476

the algebraic expression for the electric field.1484

I'm going to go to Gauss’s law again, integral/ the closed surface of E ⋅ DA = Q enclosed/ ε₀.1486

The left hand side becomes E × that area of our Gaussian surface 4 π R² 1498

which I’m going to pick outside all of that 0.2 m must equal QI + QO/ ε₀.1503

Therefore, the electric field strength is QI + QO/ 4 π ε₀ R².1512

Excellent, moving on to part C, determine an algebraic expression for the potential when you are out in that same region as part B.1527

Since we already did the work to the electric field, let us use that to help solve this.1537

It is V = the opposite of the integral of E ⋅ DL is going to be the opposite of the integral 1541

coming from infinity to some point R our electric field of what we have previously QI + QO/ 4 π ε₀ R² DR.1548

We will pull out our constants -QI + QO, those are not planning on changing any time.1566

4 π ε₀ is a constant, we can pull that out, - the integral from infinity to R of R⁻² DR 1575

which implies then that our potential is - QI + QO/ 4 π ε₀ × -1/ R from infinity to R, is equal to - QI + QO/ 4 π ε₀ × -1/ R.1588

The infinity piece is going to go to 0 so we end up with QI + QO/ 4 π ε₀ R.1619

You may have been able to predict without having gone through all the math, but still good practice for you.1631

Part B, using the numerical information given, calculate the value of the total charge QT on the 2 spherical shells which is QI + QO.1638

Since we know the potential there, V = our total Q, QI + QO/ 4 π ε₀ R and we know R = 0.2 m, 1649

when V = 100 v, that implies then that QT = 4 π ε₀ R × our potential.1666

Or 4 π ε₀ × 0.2 × our 100 volts or about 2.23 × 10⁻⁹ C, or 2.23 N/C.1677

There is a part E here, on the axis, sketch the electric field E as a function of R, that the positive direction be radially outward.1697

Alright, let us see if we can sketch these in.1706

We have got a graph here and we are going to have another one here in a minute.1710

Let us make both at the same time.1715

There is our electric field and in a minute we are going to be asked about potential.1728

Let us put in our markers for 0.1, 0.2, and 0.3 m.1739

Sketch the electric field as a function of R.1746

In our first situation there is 0.1, in between 0.1 and 0.2 where QI is less than 0, we got something like that.1749

Between 0.2 onward, we are going to be coming down like that.1763

For potential in part F, we are asked to do something similar.1772

Sketch the electric potential as a function of R.1775

Let us put in our marks again here for 0.1, 0.2, and 0.3.1778

It is important to note that the negative of the slope and potential is going to give us the value for electric field.1786

That will help us plot this.1792

As we go through here, the negative slope of that gives us this.1793

If we have got a 0 here, the slope down here has to be 0 and we know that we are going to start off here at -100 V.1798

Then up here at 0.2, we are going to have to come up to + 100 V.1807

It looks like we go from a steep slope to a lower and lower slope.1814

This is probably going to look something like that.1821

And then from here on, we have got a shape something like that.1825

That the opposite of the slope of this graph gives us the value of the electric field graph.1831

Alright let us take a look at the second example.1839

This one, let us pull from the 2010 APC E and M exam free response question number 1.1842

This is one of the more challenging questions that I have seen especially as you are just learning about potential.1848

Let us take a look here and it gives us a charge + Q/ 1/4 circle of radius R but a couple of points there.1856

It says rank the magnitude of the electric potential of those points from greatest to least and justify those rankings.1865

I would say that the potential at B must be greater than the potential at A and the potential at C, which are the same.1872

You can see A and C are the same by symmetry.1880

The reason being is B is closest to the charge, therefore it must have the highest potential.1883

A and C of the same potential by symmetry, and their lower potential in B because they are further away from the charges.1889

Explain that somehow in your answer as you justify that.1895

For part B, let me find where is B?1900

We have adjusted the problem a little bit.1904

Determine an expression for the electric potential at point P due to the charge Q.1909

The P is right in the center there due to the charge Q.1916

B, we can find the potential there that is going to be the sum over all those little tiny bits of 1/ 4 π ε₀ charged from each of those bits, 1920

÷ the distance from each of those bits which should be constant.1932

That is going to be 1/ 4 π ε₀ integral from, let us go with θ = 3 π/ 4.1936

That θ = 5 π/ 4 as we divvy up the pieces of that circle.1946

Our Q is going to be the linear charge density × the radius × the differential of θ as we go around that.1952

I can draw a picture here that might help explain where we would get that from.1960

If we have our charges there going to go from 3 π/ 4 to 5 π/ 4.1966

If we break this up into little tiny pieces, there is our point P.1974

That distance we will call our R and the charge enclosed there, that Δ Q 1980

is going to be linear charged density λ × the radius × D θ, where D θ is the angle as we are going through there.1987

Θ going from 3 π/ 5 π/ 4 to 3 π/ 4.1999

Of course, λ is going to be charged / our length which is going to be total charge / ¼ circle 2007

is going to have a length of 2 π R ÷ 4, which is going to be π R ÷ 2, which is 2 Q/ π R.2015

We can continue this by saying that this then is our DQ λ R D θ/ R, which implies then that potential is going to be λ, 2027

that should be constant, we can pull that out, / 4 π ε₀ integral from 3 π/ 4 to 5 π/ 4 around the circle of the D θ, 2038

which is going to be λ/ 4 π ε₀ as we go from 3 π/ 4 to 5 π/ 4, that is π/ 2.2058

But going back to our definition of λ, we know that λ = 2Q/ π R.2071

That is going to imply then that our potential = λ 2Q/ π R.2082

We have a π in the numerator and we still have a 4 π ε₀ in the denominator.2094

Λ 2Q/ π R, 4 π and 4 π ε₀.2104

We got a 2 down here as well which is going to give us, when we do some simplifications Q/, let us see what do we have.2108

4 π ε₀ R.2118

Not bad, the map on that one once you get it setup.2124

Alright let us take a look there now at part C.2128

For C, it says we have a positive point charge Q with mass M placed at point B and released from rest.2137

Find an expression for the speed of the point charge and it is very far from the origin, so after it is gone a long ways.2143

This looks like a great spot again, these conservation of energy where the initial electric potential energy 2149

is going to be equal to the final kinetic energy when it is a long ways away.2156

Therefore, Q × the potential at point B is going to be equal to ½ Mb velocity².2161

We already found the potential at P, we said VP is 1/ 4 π ε₀ Q/ R.2171

We can then say that we have Q × VP which is Q/ 4 π ε₀ R equal to ½ Mb².2183

Rearranging this to get V all by itself, we can say that V² = 2q Q/ 4 π ε₀ MR.2202

A little bit of simplification here that becomes a 2.2220

Finally, solving for V itself, we get q Q/ 2 π ε₀ MR √.2226

The velocity is your long ways away, converting that electric potential energy into kinetic energy.2241

We got a part D on this question.2248

We are given a ⋅ on a couple of axis that looks like this.2251

Indicate the direction of the electric field at point P due to the charge Q.2265

The direction of the electric field at point P should be to the right.2271

Do not have any vertical component due to symmetry, that should be pretty straightforward2279

Finally, part E.2285

Let us go to blue here.2290

Derive an expression for the magnitude of the electric field at point P.2293

We are only worried about the X direction by symmetry.2297

The electric field is just the X component of the electric field, that will be the integral of all the little pieces of electric field X 2300

due to the portions around there, which will be the integral of DE cos θ which is the integral of DQ/ 4 π ε₀ R² cos θ.2310

We have got to do a little bit of work to rearrange this to make it look a little bit easier.2332

We already know that λ is going to be 2Q/ π R, I think we said.2337

DQ was λ R D θ which is going to be 2Q/ π R, there is our λ R D θ.2348

Or DQ = 2Q/ π D θ.2361

This implies then that our electric field is going to be the integral of DQ which is 2Q/ π.2370

We have still a 4 π ε₀ R² in there.2382

We have our cos θ term and we still have our D θ from DQ.2389

A little bit more work here still.2401

E is going to be equal to, let us pull out our constants, 2 and the 4 that is going to be a 2 in the denominator 2404

so we end up with Q/ 2 ε₀ π² R² integral from -π/ 4 to π/ 4 cos D θ.2411

That is a little bit more reasonable.2430

Which is Q/ 2 ε₀ π² R² the integral of the cos is the sin.2434

That will be the sin of π/ 4 - the sin of -π/ 4, which implies that the electric field 2444

is going to be Q/ 2 ε₀ π² R² sin of π/ 4 √ 2/ 2 sin - the sin of -π/ 4 is going to be + √ 2/ 2.2456

I can factor out that √ 2 and say that we get Q √ 2/, we have 2 ε₀ π² R² or Q/ √ 2 ε₀ π² R².2474

It was a little bit of tricky but certainly doable especially when you get to the point where you have just the cos θ D θ for your integral.2501

Let us do say, to think we have 2 more of these to get through.2511

Let us try the 2009 exam free response number 1.2515

In this one we are given a spherically symmetric charged distribution with net positive charge Q0 distributed within some radius R.2525

Its electric potential as a function of distance is given by those formulas.2535

That is kind of complicated formulas there.2541

For the following regions, indicate the direction of the electric field ER and derive an expression for its magnitude.2544

For A1, we are looking inside that sphere.2553

The way I would start this is if we are given the potential, the electric field is - DV DR 2558

which is - the derivative with respect to r of R function for V which is going to be Q₀/ 4 π ε₀ R is given to us, × -2 + 3 × r / R².2566

Which is, we can pull out our constants -Q0/ 4 π ε₀ RD/ DR -2 + 3 × R/ R², 2585

which implies then that our electric field = –Q0/ 4 π ε₀ R ×, our derivative here is going to be 6R/ R².2603

Putting all that together, -3Q₀ r/ 2 π ε₀ R³, complies that the magnitude of the electric field would be 3Q₀ R/ 2 π ε₀ R³.2623

That is going to be inward since E is less than 0.2645

For part A2, we want to do that for r greater than R.2654

Using the same strategy, E = - DV DR is going to be -D/ DR of Q₀/ 4 π ε₀ R.2660

Pull out our constants, -Q/ 4 π ε₀ × the derivative with respect to r of R⁻¹.2674

Q is just going to be Q/ 4 π ε₀ R².2684

No surprise there, we can treat it as if it was a point particle at the center of the sphere.2692

Of course, that is going to be outward since E is greater than 0.2697

Part A, check.2703

Alright part B, for the following regions, derive an expression for the enclosed charge that generates the electric field in that region.2705

The first one is for inside that sphere r less than R.2713

I'm going to go to Gauss’s law here and say that the integral over the close surface of E ⋅ DA is our enclosed charge ÷ ε₀, 2717

complies then the left hand side E × the area of our Gaussian sphere 4 π R² = Q enclosed/ ε₀,2734

which implies then that Q enclosed is going to be ε₀ E × 4 π R².2744

Now since r is less than R, we know the electric field, we just solved for that up here in black, E is -3Q₀ R/ 2 π ε₀ R³.2755

We can substitute that in to find Q enclosed = ε₀ ×, we have got our -3Q₀ R ÷ 2 π ε₀ R³.2772

We still have our 4 π R².2797

Just simplifying this, canceling, and clean this up a little bit.2803

Q enclosed would be, we have got the 3 there, ε₀, all of that should give us -6 Q₀ R³/ R³.2806

I think we are good.2829

Let us highlight that.2833

For B2, they want us to do this outside that sphere.2836

Let us give ourselves more room here.2843

The integral/ the close surface of E ⋅ DA is Q enclosed/ ε₀ which implies that, 2846

we got 4 π R² ε₀ E = Q enclosed, just like we had from the last part of the problem but substituting in our values now.2858

Knowing here that we have our electric field equal to Q₀/ 4 π ε₀ r², 2868

we can state that our left hand side becomes 4 π R² ε₀ × our electric field portion Q0/ 4 π ε₀ r².2879

All of that = Q enclosed, which implies then that all of this cancel out, cancels out the Q enclosed = Q₀, 2900

which is kind of it has to, the enclosed charge Q₀ are going to have to be the same there.2910

For part C, is there any charge on the surface of the sphere?2919

Let us answer that and we are going to justify it.2924

Let us start by saying, yes there is charge on the surface, there has to be and then we will justify and prove it.2927

How do we know that?2941

For r greater than R, the charge enclosed that r greater than R is the surface charge + whatever is enclosed at r = R.2942

This is at r greater than R, which implies then that the surface charge = the enclosed charge 2957

- the enclosed charge and we calculate that at r = R.2969

There is a difference in those two so the surface charge = Q enclosed for r greater than R,2975

- what we had for charge enclosed to an r is equal the R, which we said was Q₀ - -6Q₀ R³/ R³.2984

Which is Q₀ - -6Q₀ or 7Q₀.3000

It asks us to do some graphing.3012

Let us go to a new page here for the graphing and draw our axis here first.3015

We are asked to graph a force that would act on a positive test charge in the region r less than r and r greater than r, 3031

assuming that the force direct radially outward is positive.3039

There is r and let us make a mark here for R.3043

For r less than r, we know that we are going to have something that is proportional to r.3052

But since we are saying they we are calling radially outward positive, it is going to look something like that.3058

R is proportional to r.3067

Once we are outside of that, we get to our more familiar portion where we are proportional to 1/ R².3072

1 or 2 more example problems here.3096

Let us take a look at the 2007 APC E and M exam free response question number 2.3102

We are given a figure with a non conducting solid sphere of radius A and charge + Q uniformly distributed throughout its volume 3111

and we also have a non conducting spherical shell with radius 2A and out radius 3A with a charge -Q uniformly distributed through its volume.3120

We are asked to start off with Gauss’s law.3129

Looking at A1 here, use Gauss’s law to find the electric field within the solid sphere.3133

Let us start off with the volume charge density being Q/ V, it is uniformly distributed it said.3141

That is going to be Q/ 4/3 π A² or 3Q/ 4 π A³.3146

When we apply Gauss’s law, the integral/ the close surface of E ⋅ DA is equal to the enclosed charge ÷ ε₀, 3160

which implies then that the electric field × our Gaussian sphere 4 π R² area of our Gaussian sphere.3171

It is going to be equal to our volume charged density × the volume that we are enclosing in our Gaussian sphere 4/ 3 π r³/ ε₀.3179

We can substitute in for ρ where we know that ρ is 3Q/ 4 π A³, 3196

to state that the electric field then is going to be, we got our ρ here 3Q/ 4 π A³.3207

We have our 4/3 π R³, we have got an ε₀ down here and we are also dividing by 4 π R² from the left hand side.3221

All of that is going to simplify to, we have got a 3 and 3, we have got a 4 and 4, we have got a π and a π.3232

We got and R² and R³, I would say then that E = Qr/ 4 π ε₀ A³.3247

For part 2, it wants us to find the electric field between the solid sphere and the shell.3267

Between those two, this is probably going to be a little bit simpler as we have gotten in the Gauss’s law.3272

A2 integral/ the close surface of E ⋅ DA = the enclosed charged ÷ ε₀, complies that E × 4 π R² = that enclosed charge/ ε₀ Q.3279

Therefore, E is just equal to Q/ 4 π ε₀ R².3299

Even if you can do that from your head, notice it tells you to use Gauss’s law to derive the expression.3308

Take the time to write all of that out.3313

Alright A3, let us give ourselves more room to do a new picture.3317

Here we have, within the spherical shell 2A to 3A.3330

Q enclosed = Q + ρ B × VB, where ρ B is -Q/ VB, which is going to be -Q/ 4/3 π × 3A³ -4/3 π × 2A³,3337

which is going to be equal to -Q/ 4/3 π 27 A³ -8 A³.3369

Ρ B = -Q/ 4/3 π 19 A³.3382

Applying Gauss’s law then, integral over the closed surface of E ⋅ DA = Q enclosed/ ε₀.3396

Therefore, the electric field 4 π R² ε₀ equal to, we have got Q + -Q/ 4/3 π 19 A³ multiplied by 4/ 3 π R³ -4/3 π 2A³.3406

Which implies then that the electric field, our left hand side × 4 π ε₀ R² = Q × 1 – R³ / 19 A³ + 8 A³/19 A³.3442

That is the algebra coming up here looks like.3467

Q × 1 – R³/ 19 A³ + 8/ 19, which implies then that the electric field = Q/ 4 π ε₀ R² × 1 – R³/ 19 A³ + 8 A³/ 19 A³, it is equal to Q/, factoring out that 19.3470

4 × 19 = 76 × π × ε₀ × R², 19 – R³/ A³ + 8, which implies that the electric field = Q/76 π ε₀ R² × 27 – R³/ A³.3511

Quite the problem going on there.3543

A4, outside the spherical shell where r is greater than 3A, Q enclosed = 0.3553

Therefore, E = 0.3569

Alright, that one was not so bad.3572

For part B, what is the electric potential of the outer surface of the spherical shell where R = 3 A and explain your reasoning.3574

Let us solve, E = - DV DR which implies then that V = - the integral of E DR, 3583

which is - the integral from infinity to 3A of E ⋅ DR which is - the integral of 0 or 0.3594

We will justify it there.3606

C, derive an expression for the electric potential difference between points X and Y as shown in the figure.3609

For part C, VX - VY which is Δ V is the integral from A to 2A of E ⋅ DR, 3619

which is the integral from A to 2A of Q/ 4 π ε₀ R² DR.3631

Pulling out our constants, Q/ 4 π ε₀ integral from A to 2A of R⁻² DR.3642

Or Q/ 4 π ε₀ the integral of R⁻² is going to be -1/ R evaluated from A to 2A.3654

It can be Q/ 4 π ε₀ × -1/2 A - -1/ A or we can say then that we have Q/ 8 π A ε₀.3669