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Course Learning Assistance Physics Learning Center (Monday Wednesday 2:00-4:30and 6:00 to 8:30 in rooms 129 and 130 Physics) LEAD/tutoring sessions (will start in 2nd/3rd week, ss.mst.edu/tutoring/ Disability Support Services (accommodation letters)http://dss.mst.edu/ Testing Centerhttp://testcenter.mst.edu/

Exam 1Exam 1 is on Tuesday, February 14, from 5:00-6:00 PM.According to the Student Academic Regulations “The period from 5:00 – 6:00 PMdaily [is] to be designated for common exams. If a class or other requiredacademic activity is scheduled during common exam time, the instructor of theclass that conflicts with the common exam will provide accommodations for thestudents taking the common exam.” other time conflicts are the student’s responsibilityunless the student participates in a major university orintercollegiate event

Career FairThe spring career fair is on Tuesday, February 21, 2017. Thereis no conflict between the exam and career fair.If you have an interview or must attend a career fair activityduring your recitation time on February 21, contact yourrecitation instructor.We understand the importance of the career fair andwill work with you to make sure you can participate!

LabsLabs begin week of Jan ��Odd” labs (3L01, 3L03, etc.) start the week of January 23“Even” labs (3L02, 3L04, etc.) start the week of January 30Purchase a lab manual in the Physics office. Students notpurchasing a lab manual will receive a lab grade of zero.Lab manuals are not available at the bookstore.

Review of Lecture 1, Part ICoulomb's Law:1 q1q 2F ,212 4πε 0 r12r12 -q1q2Coulomb’s Law quantifies the force between charged particles.

Review of Lecture 1, Part IIElectric field: charges create electric fields electric field of a point charge:qmagnitude: E k 2rdirection: away from ,towards force felt by charge in electric field:-F qEpositive charge feels force in direction of Enegative charge feels force in direction opposite to E

Today’s agenda:Electric field due to a charge distribution.You must be able to calculate electric field of a continuous distribution of charge. ---

Electric FieldDue to a Continuous Charge DistributionProblem: our equations for the Coulomb force and the electricfield hold for point charges onlySolution: decompose extended object into charge elements calculate electric field for each element sum up (integrate) contributions of all elements toobtain total electric field

Electric FieldDue to a Continuous Charge Distribution(worked examples)finite line of chargegeneral derivation: http://www.youtube.com/watch?v WmZ3G2DWHlgring of chargedisc of chargeinfinite sheet of chargeinfinite line of chargesemicircle of charge

Instead of talking about electric fields of charge distributions,let’s work some examples. We’ll start with a “line” of charge.Example: A rod* of length L has a uniformly distributed totalpositive charge Q. Calculate the electric field at a point Plocated a distance d below the rod, along an axis through theleft end of the rod and perpendicular to the rod.Example: A rod* of length L has a uniformly distributed totalnegative charge -Q. Calculate the electric field at a point Plocated a distance d below the rod, along an axis through thecenter of and perpendicular to the rod.I will work one of the above examples at the board in lecture.You should try the other for yourself.*Assume the rod has negligible thickness.

Example: A rod of length L has a uniformly distributed totalnegative charge -Q. Calculate the electric field at a point Plocated a distance d below the rod, along an axis through thecenter of and perpendicular to the rod.-QLqStarting equation: E k 2rdP“Legal” version of startingequation:dqdE k 2rThis is “better” because it tells you howto work the problem! It also helps youavoid common vector mistakes.

You should begin electric field of charge distribution problemswith thisdqdE k 2rThis is a “legal” version of astarting equation, so it is “official.”because the equation “tells” you how to work the problem.The equation says:(1) pick a dq of charge somewhere in the distribution(2) draw in your diagram the dE due to that dq(3) draw the components of dE(4) for each component, check for simplifications due tosymmetry, then integrate over the charge distribution.

Calculate the electric field at a point P.y-QLdqStarting equation: dE k 2rxdq dEdPPick a dq (best to not putit at either end or in themiddle).Draw the dE due to the dq.Before I draw the components, I need to define axes!Now draw the components.Do you see why symmetry tells me that Ex 0?

Calculate the electric field at a point P.y-QLxEx 0, so calculate Eyxdq dEd PFirst, label an angle inthe vector diagram.dEy dE sin yes, the sign conveysimportant informationTo find sin , we need the x-coordinate of dq. If dq is at anarbitrary position along the x-axis, what is a good name for itscoordinate? That’s right, we’ll call it x.The diagram is getting rather “busy,” but we are almost donewith it.

Calculate the electric field at a point P.y-QLxxdq dEd PTo find sin , look at the green triangle. The sides have lengthx and d, and hypotenuse r, wherer x 2 d2From the green triangle, we see that sin d / r.

Calculate the electric field at a point P.y-QLxxdq dEd PNow we start to put things together:dqdq dd dqdqdE y dE sin k 2 sin k 2 k 3 kd22 3/2rr rr x d To find Ey we simply integrate from one end of the rod to theother (from –L/2 to L/2).

Calculate the electric field at a point P.y-QLxxdq dEd PEy L2 L 2kd dq x2 d 2 3/2But wait! We are integrating over the rod, which lies along thex-axis. Doesn’t there need to be a dx somewhere?

Calculate the electric field at a point P.y-QxdqxI removed un-needed “stuff” from the figure.dq is a tiny bit of charge on the uniformly charged rod.If the charge is uniformly distributed, then the amount ofcharge per length of rod is(charge)(linear charge density) (length)orQ L

Calculate the electric field at a point P.y-QxxdqQ LWe use the symbol for linear charge density. Youprobably thought (based on Physics 1135) that isthe symbol for wavelength. It is. But not today! charge on segment of rod charge length length of segment of rod What would be a good name for an infinitesimal length of rodthat lies along the x-axis? How about dx?

Calculate the electric field at a point P.ydx-QxxdqWe can take outside the integralbecause the charge is uniformlydistributed, so must be constant.Thus, dq dx andEy L2 L 2dqkdQE y kdL x 2 ddxL2 L 2 x 2 3/22 d L2 L 2 2 3/2kd dx x2 d 2 3/2 kd dxL2 L 2 x2 d 2 3/2The physics of the problem is alldone. The rest is “just” math.

Calculate the electric field at a point P.ydx-QxxdqQE y kdL dxL2 L 2 x2 d 2 3/2A note on the “just” math part. We expect you to remember derivativesand integrals of simple power and trig functions, as well as exponentials.The rest you can look up; on exams we will provide tables of integrals. Wewould provide you with the above integral. It is not one that I could do in5 minutes, so I would not expect you to do it.

Example: A rod of length L has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at apoint P along the axis of the rod a distance d from one end.*PdLTo be worked at the board in lecture *Assume the rod has negligible thickness.

Example: A rod of length L has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at apoint P along the axis of the rod a distance d from one end.yPxdLIt’s a good bet we will need x- and y-axes, so let’s just put themin right now. Let’s put the origin at P.After the previous example, we realize we will need to calculatethe linear charge density on the rod.Q and Q LLNote 1: both and Q are given, so we can express our answer in terms of our choice of either one.Note 2: we are told Q is positive, so we don’t need the absolute value signs as in the previous example.

Example: A rod of length L has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at apoint P along the axis of the rod a distance d from one end.ydEdqPdxLdqWe start with our usual equation: dE k 2rRemember, this is thebest way to start aproblem like this one.The equation says pick a dq of charge, so do it!Because dq is positive, its contribution to the electric fieldpoints away from the rod.

ydExdqPdxLI could work the problem “all at once” using unit vectornotation, but for now I think it’s safer to work out eachcomponent separately.The electric field at point P has no y-component (why?).Therefore, Ey 0.The rod hasno thickness.The infinitesimal charge dq is a distance x away from the origin.dq dx dx -k 2Therefore dE x -k 2 -k2xxxThe – sign isimportant!If I don’t know the sign of dq, I keep the absolute value signs but don’t know the direction of dE x. If dq is negative, then thesafest thing to do is change the – signs in the equation above to ’s, and keep the absolute value signs around dq.

ydExdqPdxLNow simply integrate over the rod.E x dE x rod d Ldd L dx dx-k 2 -k dxx2 d d L 1 1 1E x -k k k x d d L d d d L d Lk LEx d d L

ydExdqPdxLThe problem asks for the electric field at point P. Let’s make iteasy for a potential grader by writing down our completeanswer with a box around it.Any of the boxed answers below is correct.E y 0k LEx d d L k LE , in the -x directiond d L If a problem says “express your answer inunit vector notation,” you need to do that!k L ˆE id d L On an exam, put a box around each part of an answerwhen you finish it, so the grader can clearly see it. You cancopy parts of an answer and rewrite them together in oneplace so you can put a box around the whole answer atonce (like I did here), but don’t make a mistake copying,because you will lose points. Also, just box one answer. Donot box different versions of the same answer.

Example: calculate the electric field due to an infinite line ofpositive charge.There are two approaches to the mathematics of this problem.One approach is that of example 21.10. See notes here. Analternative mathematical approach is posted here. The result is 2k E 2 0 rrThis is not an “official”starting equation!The above equation is not on your OSE sheet. In general, youmay not use it as a starting equation!If a homework problem has an infinite line of charge, you wouldneed to repeat the derivation, unless I give you permission touse it.

Example: A ring of radius a has a uniform charge per unitlength and a total positive charge Q. Calculate the electric fieldat a point P along the axis of the ring at a distance x0 from itscenter.To be worked at theblackboard in lecture.Pxx0Homework hint: you must provide this derivation in your solution to anyproblems about rings of charge (e.g. 21.53 or 21.55, if assigned).Visualization here (requires Shockwave, which downloads gIntegration/RingIntegration.htm

Example: A ring of radius a has a uniform charge per unitlength and a total positive charge Q. Calculate the electric fieldat a point P along the axis of the ring at a distance x0 from itscenter.An edge-on view of the ring would look like this:yxThe z-axis would be coming out of the screen at you.I will use the perspective view of the ring in my solution.

Example: A ring of radius a has a uniform charge per unitlength and a total positive charge Q. Calculate the electric fieldat a point P along the axis of the ring at a distance x0 from itscenter.yLet’s add a y-axis to thefigure.dqra Px0 xdEStarting equation:dqdE k 2rPick a dq of charge. Let’s put it on the y-axis for now.Show the dE due to that positive dq.We’ll need r and later.

ydqradE′ x0rdq′PdExShow the x- and ycomponents of dE. Theremay also be a z-component,which we’ll leave outbecause it is difficult to drawand visualize.Consider the dq’ on the ring where it is intersected by thenegative y-axis.dq’ gives rise to dE’ at P. Show the components of dq’.All points on the ring are the same distance r from point P.Also, x0 and are the same for all points on the ring.The y-components cancel pairwise! Same for thez-components (not shown). Ey Ez 0.

yBack to our OSE dqra x0Px dE x dE cos dEFrom the diagram:dqdE k 2rr x a202x0cos rFor a given x0, r is a constantfor all points on the ring.Only works because all Ex are in same direction. dq x 0x0E x dE x k 2 k 3r rrringring kx 0 Qx0 ring d q k r 3 Q x 2 a 2 3/2 0

dq x 0x0E x dE x k 2 k 3r rrringring kx 0 Qx0 ring d q k r 3 Q x 2 a 2 3/2 0 Some of you are wondering why all the absolute value signs.You don’t really need them in this example, because Q ispositive.When I draw the dEx and dEy in the diagram, the sign of Qdetermines the directions of the components.Because I used the sign of Q to determine the directions of thecomponents in my diagram, I don’t want to accidentally use thesign again later and get the wrong direction in my final answer;hence the absolute value signs, for safety.If x0 is negative, then Ex points along the –x direction, as it should, so I don’t wantto put absolute value signs around the x0 in the answer.

yBack to our example aPx0Ex xE xkx 0 Q20 a 2 3/2E y Ez 0Also “legal” answers:E kx 0 Q2x 0 a 2 3/2ˆiE kx 0 Q2x 0 a 2 3/2, away from the centerThese equations are only valid for P along the x-axis!Awesome Youtube Derivation: http://www.youtube.com/watch?v 80mM3kSTZcE(he leaves out a factor of a in several steps, but finds it in the end).What would be different ifQ were negative? If P wereon the negative x-axis?

Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axisof the disc at a distance x0 from its center.PRx0x

Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axisof the disc at a distance x0 from its center.The disc is made ofconcentric rings.rPRxx02 rdrThe ring has infinitesimal thickness, so you can imagine it asa rectangular strip.Imagine taking a ringand cutting it so you canlay it out along a line.The length is 2 r, thethickness is dr, so thearea of a ring at a radiusr is 2 rdr.Caution! In the previous example, the radius of the ring was R. Here the radius of the disc isR, and the rings it is made of have (variable) radius r.

Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axisof the disc at a distance x0 from its center.dqThe charge on eachring is dq (2 rdr).Let’s assume is positiveso dq is positive.r charge on ring charge per area area PRx0xdEringWe previously derived an equation forthe electric field of this ring. We’ll callit dEring here, because the ring is aninfinitesimal part of the entire disc.Skip to slide 47 for result.dE ring kx 0 dq ring x20 r 2 3/2

Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axisof the disc at a distance x0 from its center.Let’s assume is positiveso dq is positive.dqdE ring rPx0RE disc dEdiscring disc kx 0 2 rdr x x20 r 2 3/2xdEring kx 0 dq ring20 r 2 3/2 kx 0 R0 xkx 0 (2 rdr ) x20 r2r dr20 r 2 3/2 2 3/2

Let’s assume is positiveso dq is positive.dqrPRx0xdEringE disc kx 0 R0 x2r dr20 r 2 3/2 x2 r 0 kx 0 1/ 2 2 1/2E discYou know how to integratethis. The integrand is just(stuff)-3/2 d(stuff)R x0 2k x 0 x 0 x 2 R 2 1/2 0 0 Kind of nasty looking, isn’t it.

PRx0xEdisc x0x0 E x 2k x 0 x 2 R 2 1/2 0 As usual, there areseveral ways to writethe answer.E y Ez 0 xx0 ˆiE 2k 0 x 0 x 2 R 2 1/2 0 Or you could give the magnitude and direction.

Example: Calculate the electric field at a distance x0 from aninfinite plane sheet with a uniform charge density .An infinite sheet is “the same as” disc of infinite radius.E sheet xx0 lim 2k 0 R x 0 x 2 R 2 1/2 0 1Take the limit and use k to get4 0E sheet .2 0This is the magnitude of E. The direction isaway from a positively-charged sheet, ortowards a negatively-charged sheet.

Example: Calculate the electric field at a distance x0 from aninfinite plane sheet with a uniform charge density .E sheet .2 0Interesting.does not depend on distance from the sheet. Doesthat make sense?This is your fourth Official Starting Equation, and the only onefrom all of today’s lecture!I’ve been Really Nice and put this on your starting equation sheet. You don’t have toderive it for your homework!

Example: calculate the electric field at “center” of semicircularline of uniformly-distributed positive charge, oriented as shown. QRyYou don’t have to follow the steps in the exactorder I present here. Just let the problem tell youwhat to. You may do things in a different order;that’s probably OK.xds To be worked at theblackboard in lecture.d RdE

Example: calculate the electric field at “center” of semicircularline of uniformly-distributed positive charge, oriented as shown. Qdqdsd yRStart with our usual OSE.dqdE k 2rPick an infinitesimal dq of charge.xdq subtends an arc length ds, and an angle d .What is the charge dq?dq charge per length of arc length of the arc dq ds

Example: calculate the electric field at “center” of semicircularline of uniformly-distributed positive charge, oriented as shown. Qdqdsd yxdE′dq′RdEDraw the dE due to the dq, andshow its components.Do you see any helpful symmetry?Pick a dq′ horizontally across the arc from dq. The xcomponents of dq and dq′ will cancel. Because of thissymmetry, Ex 0Each dEy points downward so Ey will be negative.

Example: calculate the electric field at “center” of semicircularline of uniformly-distributed positive charge, oriented as shown. Qdqdsdqy xRdE Recall that dq and ds areinfinitesimal. dq is located at anangle along the semicircle fromthe negative y-axis. is also one of the angles in thevector triangle.dE y -dE sin

Example: calculate the electric field at “center” of semicircularline of uniformly-distributed positive charge, oriented as shown. Qdqdsdqy xRdE An arc of a circle has a lengthequal to the circle radius timesthe angle subtended (in radians):ds R d Also, charge on arc Q Q length of arc 1 2 R R2

Qdqdsdqy xRdE Let’s summarize what we havedone so far.dqdE k 2rQ Rdq dsds R d Every dq is a distance R away from the arc center: Q R d dsQ d R dE k 2 k k2RR R 2dE y -dE sin r R

Qdqdsdqy kQ d dE y sin 2 RRdEx kQEy - 2 RkQ arc sin d - R 2Ey dEarcy kQ d sin 2 R arc kQ 0 sin d R 2 cos 0kQkQ2kQE y 2 cos cos 0 2 1 1 - 2 R R RAwesome Youtube derivation: http://www.youtube.com/watch?v L1n2EUvayfw

Help!We covered a lot of material in a brief time.If you want to explore a slightly different presentation of this atyour leisure, try the MIT Open Courseware site:http://ocw.mit.edu/courses/physics/.

Homework Hints (may not apply every semester)Your starting equations so far are:q1q 2F k 212r12F0E q0qE k 2rE sheet .2 0(plus Physics 2135 starting equations).This is a “legal variation” (use it for charge distributions):dqdE k 2 .rYou can remove the absolute value signs if you know that dq isalways positive.

Homework Hints (may not apply every semester)Suppose you have to evaluate this integral in you

If a class or other required academic activity is scheduled during common exam time, the instructor of the class that conflicts with the common exam will provide accommodations for the . Purchase a lab manual in the Physics office. . Labs Labs begin week of Jan 23! Coulomb's Law: -Review of Lecture 1, Part I q 1 q 2 r 12 12 2 0 12 1 qq F .

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