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
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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
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: 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 R0031
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 potentials0094
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 = 00399
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 DR0719
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 field0902
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 R²0930
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 wise1110
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 then1152
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 constants1192
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 R²1498
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 integral1541
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 DR1575
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 Q1980
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 circle2007
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 energy2149
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 X2300
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 denominator2404
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 field2444
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 DR2558
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 charge2957
- 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 volume3111
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
2 answers
Tue Mar 13, 2018 9:17 AM
Post by Kevin Fleming on March 13 at 09:05:42 AM
Prof. Fullerton,
Why is the potential inside the circle equal to the potential outside the circle? If the electric field inside the shell is assumed to be zero, then how can electric potential exist inside that region? Thanks.
2 answers
Fri Jul 7, 2017 2:18 PM
Post by Luka Kostic on July 7, 2017
Hello prof.
In the case of potential due to a sold spherical insulator, is it necessary to express volumic charge density or we can just leave ro, becouse it's a uniformly charged sphere?
1 answer
Wed Oct 19, 2016 7:40 AM
Post by Mohsin Alibrahim on October 10, 2016
Mr F
When you were finding potential due to a uniformly charged disk, why the charge density = Q/pi*R^2 rather than Q/pi*r^2 ?
1 answer
Fri Apr 8, 2016 6:23 AM
Post by Ayberk Aydin on April 8, 2016
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
Fri Apr 8, 2016 6:16 AM
Post by Ayberk Aydin on April 8, 2016
Why is the electric field between .1 and .2 meters negative in the 2012 FRQ?
1 answer
Tue Jan 19, 2016 6:56 AM
Post by Shehryar Khursheed on January 18, 2016
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
Mon Jan 18, 2016 1:31 PM
Post by Shehryar Khursheed on January 18, 2016
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!