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 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 straightforward*2279

*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

2 answers

Last reply by: Professor Dan Fullerton

Fri Jul 7, 2017 2:18 PM

Post by Luka Kostic on July 7 at 11:56:10 AM

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

Last reply by: Professor Dan Fullerton

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

Last reply by: Professor Dan Fullerton

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

Last reply by: Professor Dan Fullerton

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

Last reply by: Professor Dan Fullerton

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

Last reply by: Professor Dan Fullerton

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!