WebTA signing off

Well, I've had an interesting semester experimenting with this WebTA project. I have to say I missed having contact with you the students in tutorials, but it's been interesting to get your questions and challenging to try to answer them effectively over the web. Watching the traffic on the site it seems many of you are using the material I've been providing. I hope this means you have found all this helpful and have learned something from what I've posted here. I'll take this last opportunity to remind you that you can still provide feedback by filling out my survey and returning it to my mailbox (see post WebTA survey), or by emailing me or commenting here.

I wish you all the best with your courses and exams. Try to stay clear-headed for exams. Get sleep. Breath deep. Give yourself the chance to think on the exam... very often it is more effective that those last couple of hours of cramming.

Good bye and Good luck!

Sarah

Question: Max current for an electromagnet (Ch 20 #80)

I am confused about Ch 20 #80. I know Power is the rate of work and I can use that to calculate current . I would also used the magnetic field equation B = N (uI)/2r to calculate the number of turns required to run power at max power (i.e use the calculated Imax). But I don't know B and I am not sure how to calculate that from V.

Ahh... ok, it took me a minute to figure this one out, but the trick is that it is using things you know from previous chapters (tricky folks these text book writers are). Do not be distracted by the equation they give you immediately! Think back to when you looked at electrical power:

P=IV

So, you know V, and Pmax... you can directly determine Imax from this. Then you can use the equation they give to determine the maximum magnetic field that can be generated.

Oftentimes half the trouble in figuring out a problem is determing what information you need to use and when. Most of the problems in textbook are divided up into the appropriate sections where you learn the material so it is easy to determine how to approach a given "type" of problem. The trouble is in the real world there are no "types" of problems. You have to take each situation and consider what factors are present and relavent. Every now and again textbooks (and exam writers!) throw in questions that test your ability to discern what is needed to help train you to solve problems more generally. Learning how to do this takes some experience (which unfortuantely just requires time and practice), and requires that you think about the concepts and how they apply, not just classify the problem as a particular type.

Hope that helps you out.

Question: Two resistors, Two capacitors (Ch 19 #50)

Ch 19 #50 part b

I am not sure how to distinguish whether a capacitor is charging or discharging from this problem. Sometimes, I can't tell through which resistor will a capacitor will discharge...is it through the one in parallel or in series?

Let's just look at simplifying what we have when the switch is open, then closed.

When the switch is open there are two resistors in series which are in parallel with two capacitors in series. The voltage across the set of resistors will be the same as the voltage across the set of capacitors (24 V in this case). At point "a" there will be some voltage/potential between 0 and 24 V due to the voltage dropped (IR) across one of the two resistors. Likewise at point "b" there will be some voltage/potential between 0 and 24 V as there is some voltage dropped (Q/C) across one of the two capacitors. (Note that to determine this you will have to use the equivalent resistances and capacitances of each branch of the circuit to determine the net current and charge, respectively.)

Now, when the switch is closed, charge will flow from a to b or vice versa because the potential at a and b were not equal initially. Once the switch is closed this potential difference cannot be maintained. Without working through to see which is greater and thus which direction the charge will flow I can't say whether the capacitors are charging or discharging. However, I can tell you if the potential at "a" is greater than at "b" the current (charge) will flow toward the capacitors and thus the capacitors will charge (partially). Conversely if the potential at "b" is greater than at "a" the current (charge) will flow away from the capacitors and the capacitors will discharge (partially).

Good luck. This is a tricky problem.

Correction: Example Ch 16 #61

Sorry, there is an error in the example posted on the main page: Electrostatics / Ch. 16 #61

In the first part, I forgot to square the velocity, so instead of 5.4e-9 m, I should have gotten a distance of 0.115 m. The error is also carried forward into part b, instead of 1e-12 s, I should have gotten 2e-8 s for the time.

I will try to fix the original page soon.

WebTA survey

To get some feedback on the usefulness and usage of the webTA project I have developed a short survey. Paper copies should have been (will be?) available in class at some point, however, if you missed your chance to have your say and are interested in providing feedback I will give you another chance here.

You may download a PDF version of the 2006 webTA project survey HERE [35 kB PDF].

Print it out, fill it out, and return it to my mailbox (Sarah Burke, in the Grad students section) in the mailroom of the Rutherford Physics building.

(Location of the mailroom: Enter Rutherford through main doors, turn right and go through a big brown door. At the end of the hall is a staff lounge. The mailroom is on the right just before this. It's a room with a photocopier and a bunch of labelled mailboxes. Mine's on the left hand side near the beginning as they are more or less alphabetical.)

Of course any creative ways you think of to send it back electronically are good too... sorry I don't have a way to make a form-fillable pdf. :)

End of term looming...

Just a quick warning in case you were not already aware. The webTA site will not be maintained after the end of classes. I will do my best to answer as many of the questions you ask as possible until then.

Question: Stable and unstable equilibria

How do we tell the stable and unstable equilibrium mathematically? cuz using the definition in the text to determine what equilibrium i'm dealing with is ambiguous and obscure...

This is a very good question which is actually very general. Let's start with some definitions...

Equilibrium is a state of a system in which the variables which describe the system are not changing (note that a system can be in a dynamic equilibrium where things might be moving or changing, but some variable(s) which describe the system as a whole is(are) constant). One example you are all familiar with is a mechanical system in equilibium where positions of objects are not changing (ie. no net forces acting).

In a Stable equilibrium if a small perturbation away from equilibrium is applied, the system will return itself to the equilibrium state. A good example of this is a pendulum hanging straight down. If you nudge the pendulum slightly, it will experience a force back towards the equilibrium position. It may oscillate around the equilibrium position for a bit, but it will return to its equilibrium position.

In an Unstable equilibrium if a small perturbation away from equilibrium is applied, the system will move farther away from its equilibrium state. A good example of this is a pencil balanced on it's end. If you nudge the pencil slightly, it will experience a force moving it away from equilibrium. It will simply fall to lying flat on a surface.

Ok, now that we know more about equilibria it will be easier to determine what "kind" we have. Strictly speaking, mathematically we determine whether a mechanical equilibrium is stable or unstable by looking at the second derivative of the energy with respect to the coordinate of interest. If the equilibrium is at a minimum (second derivative is positive) the system is in a stable equilibrium. If the equilibrium is at a maximum (second derivative is negative) the system is in an unstable equilibrium.

However, there is a simpler way to quickly test whether an equilibrium is stable or unstable. If we think about the definitions, each involves the response to a small perturbation. So let's "apply" a perturbation. If we say change the position slightly, is there a net force? in what direction? does the perturbation result in a force which drives the system back toward equilibrium (stable) or away from equilibrium (unstable)?

If you can't easily picture the situation in your head (as with the pencil and the pendulum), what this means in practical terms is that you re-calculate the forces at some position near, but not at equilibrium and determine whether they are driving the system back towards equilibrium (again, this means stable) or away from equilibrium (this means unstable).

Hope that clarifies things!

Question: pulling apart capacitor plates

A parallel-plate capacitor with plate area 2.0 cm^2, air-gap .5mm, is connected to a 12V battery, if the air-gap is pulled to .75mm, how much work is required?

calculated Q=4.2E-11 C, V(d=0.75)=18V, delta PE=Q*delta V = 2.52E-10 J, but the answer is 1.3E-10 J, which could be a half of my answer. PE/2? why? Would you please explain this to me? Thanks

Ok, so the thing you are changing when you pull a parallel plate capacitor apart is the capacitance (it's ability to store energy).

The capacitance depends on distance like:

C=εo*(A/d)

so clearly the capacitance will decrease if the distance between the plates is increased. Let's consider the capacitance before (C), and after (C'):

C=εo*(A/d)
C'=εo*(A/d')

Ok. So now let's consider how this will effect the energy stored in the capacitor. This can be written in 3 different ways (by substituting Q=CV into the first):

PE=(1/2)QV=(1/2)CV²=(1/2)Q²/C

Which of these should we use? Well, since the plates are connected to a 12 V battery, the voltage will be the same before and after moving the plates. Note that the charge may change so we can't use the expressions with Q in them because we don't know how Q is changing (ok, yes, we could figure it out, but it would be equivalent to using the second expression, so let's not bother).

All right, so let's put it together:

W=-ΔPE=-((1/2)C'V²-(1/2)CV²)
W=-((1/2)εo*(A/d')V²-(1/2)εo*(A/d)V²)
W=-εo*A*V²/2((1/d')-(1/d))

Plugging in the numbers given in the equation above for the work done, I get: 8.49e-11J (using 12V). You also write 18V in your question, with that number I get an answer: 1.91e-10J. Sorry to muddy the waters with more answers... the method should be correct. (maybe you can recheck which numbers the question uses?)

If you haven't already, I'd suggest taking a look at the similar example (though with constant charge not voltage) posted on the main page: Ch. 17, #50.

Hope that helps. Feel free to ask questions by posting comments.

WebTA traffic

Curious about how many of your cohorts are using the WebTA site? Here are the stats from most of the term:

Main site traffic by week:





Blog site traffic by week:

Fair Game! Topic open: everything

At this point in the year, everything is fair game for you, so any question you have is fair game for me. Feel free to ask me anything you have covered in the course (or even things from last semester that might come up).

On that note, a piece of wisdom from my 3rd year mechanics professor:

You don't have problems, you have exercises! ~Dr. Melvin Calkin

Question: difference between RC, RL, LC

How do you tell the difference between RC, RL, LC circuits?

Each of these circuits, and what they qualitatively do in DC and AC circuits...

RC - a resistor and capacitor in series. Exhibits charging behaviour with characterisitic time constant with DC voltage source. Acts as a high pass filter (allows high frequency currents, but not low frequency currents) in AC circuits.
RL - a resistor and inductor in series. Acts as a short with a DC voltage source, but smooths out rapid variations in current. Acts as a low pass filter (allows low frequency currents, but not high frequency currents) in AC circuits.
LC (and RLC) - an inductor and capacitor (and resistor) in series. If initially charged, has oscillitory behaviour (damped if also has a resistor). Has resonant behaviour with AC driving voltage (damped if also has a resistor).

Question: LC in parallel

How to calculate the resistance of a capacitor and an inductor connected in parallel?
(-XcXL/(XL-Xc))^2 ?

This is a little bit trickier that considering elements in series because the inverse of the impedances have to be added. I'm pretty sure it is beyond the scope of this course to need to do this calculation, but I will put it here anyway. I'm also not sure how you would arrive at the answer without using the appropriate complex represenations of the impedance, so I will use them but try to be careful to explain.

So, for elements in parallel:

1/Z_tot=1/Z_C+ 1/Z_L+ 1/Z_R ...[1]

We will consider just an LC circuit (with ideal components), so R=0. Now, in complex notation Z_Cand Z_L are:

Z_C=1/iωC ...[2]
Z_L=iωL ...[3]

So substituting [2,3] into [1]:

1/Z_tot=iωC+1/iωL

or, using the fact that 1/i=-i:

1/Z_tot=iωC-i/ωL
1/Z_tot=i(ωC-1/ωL) ...[4]

so, we can now replace 1/ωC with X_C and ωL with X_L:

1/Z_tot=i(1/X_C-1/X_L) ...[5]

Now we can simplify the part in the brackets, giving us:

1/Z_tot=i(X_L-X_C)/X_CX_L ...[6]

or taking the inverse of the fraction on both sides (and once again using 1/i=-i):

Z_tot=-iX_CX_L/(X_L-X_C) ...[7]

Now, if we had also had a resistor in parallel in the circuit (R), we would end up at equation [7] with both real and imaginary parts. To find X_tot we would have to find the magnitude of the resultant phasor. Since Z_tot is purely imaginary, we need only to find the magnitude of this number, which we can express as:

X_tot=-X_CX_L/(X_L-X_C) ...[8]

Which is pretty much what I think you suggested. Note that if we hadn't used the imaginary numbers we would not have been able to get the right signs and would have come up with a different answer.

Looking back at eqn. [7], let's think about what this represents... In the imaginary plane Z_tot would be a phasor pointing along the y-axis. For values of X_L and X_C which make Z_tot positive, the impedance could be considered "inductive", and inversely if Z_tot were negative, the impedance could be considered "capacitive".

Let's consider two extreme cases...

If the frequency is high, X_L is very large, and X_C is very small (let's say negligible). In the numerator of Z_tot the frequency cancels out, so this is a constant. If we look at the denomenator, X_L >> X_C, so the denomenator is approximately X_L. Since X_L is large, Z_tot will be small and negative (capacitive). In fact if you substitute in equations [2,3] ignoring X_C in the denominator, you will recover Z_C. This is consistant with the inductor providing a break in the circuit, and the capacitor providing a short.

Similarly, if the frequency is very low, X_C becomes very large, and X_L very small. In this case, the result is small and positive (inductive), and indeed again, if you work it out you will recover Z_L as Z_tot. This is consistant with the capacitor providing a break in the circuit, and the inductor providing a short.

To summarize:

ω→∞ Z_tot→Z_C and current flows through the capacitor
ω→0 Z_tot→Z_L and current flows through the inductor

These limiting cases can be worked out simply as a thought experiment (consider which branch of the circuit has a small resistance and which has a large resistance) and are often more insightful than getting through the algebra.

Hope this helps.

Topic Open: RC, RL, LC, LRC circuits

The following topic is now open for questions: RC, RL, LC, LRC circuits (and any AC circuits)

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Impedance of LR circuit with "real" inductor

Just had a question about LR circuits. Just say there was a circuit consisting of a resistor and inductor. And just say the inductor had a resistance of its own. How do you find the total impedance of the circuit. This was similar to one of the CAPA questions. My friend told me to find the impedance of just the resistor and inductor and then add the resistance of the inductor to Z. I am confused on why you do this? Why couldn't you just add the resistance of the inductor and resistor first, then use the formula Z = sqrt(R^2 + XL ^2), R being the resistance of the inductor and resistor combined?

Actually, you are right! You can (and should) just add the resistance of the inductor to the resistance of the resistor then find the impedance including the inductive term.

Here's how it works. A real inductor will have some inherent resistance due to the fact that it is a (usually fairly large) coil of wire. To analyze a circuit, we would replace a "real" inductor with and "ideal" inductor and a resistor (to represent the pure resistance of the inductor). So, in an LR circuit you would go from having 1 resistor and 1 "real" inductor in series to having 2 resistors in series and one "ideal" inductor also in series. You would then find the net resistance of the circuit, and this would be "R" in the formula you quote above.

From this you would find the magnitude of the inductance from Z=sqrt(R²+X_L²)
(this formula comes from the fact that R and X_L are 90^o out of phase, so the magnitude is the hypotenuse of the two "phasors" see: Basics of AC circuits/RLC circuit example). Since R and X_L do not add linearly, I do not think you should get the right answer if you go about it in the opposite order (except maybe in some special circumstance).

EDIT: A similar question was also asked regarding an LRC circuit. In this case, the methodology is the same. The inductor with a resistance is replaced by a resistor and an inductor in series. The resistance of the whole cicuit is calculated, and then the magnitude of the impedance is determined. In the case of an LRC circuit, Z is given by:

Z=sqrt(R²+X_L²+X_C²)

Otherwise, the whole problem is the same.

Hope this helps!

Question: Solenoid in a coil

If a solenoid is placed inside a coil, how does the current flowing through it affect said coil? How would you find the inductance value for the coil?

That's a very good question. Qualitatively, the solenoid (with some current running through it) establishes a magnetic field parallel to the axis of the solenoid. If a coil is then placed on the outside of it, the coil will "sense" the magnetic field established by the solenoid. If the current in the solenoid is time-varying (changing in time) then the magnetic field established by it will also change with time. The coil will then "sense" this changing magnetic field and respond to oppose the change (Lenz's law).



Quantitatively, we will have to look at what the magnitude of the magnetic field produced by the solenoid is, and how that may depend on time. The magnetic field of a solenoid is given by:

B_sol=μ₀I[N/l]

Thus, if the current, I is a function of time, the magnetic field will have the same time dependence. If, say, the time dependence is linear (eg. we ramp up the current to the solenoid at a constant rate), then we can use the slope as ΔB/Δt, which is related to the EMF:

EMF=-ΔΦ_B/Δt=-A_{perpendicular}ΔB/Δt

Chapter 21, Question #80 is an example of this, and might be good to look at.

I hope this is helpful and answers your question.

Topic open: Charges and currents in magnetic fields

The following topic is now open for questions: Charges and currents in magnetic fields.

To pose a question, please post a comment to this post by clicking on the comments link below.

Topic open: Magnetic fields

The following topic is now open for questions: Magnetic fields.

To pose a question, please post a comment to this post by clicking on the comments link below.

Status check

Just wanted to check in with everyone using the webTA site. I have a couple of questions for you!

1. Am I keeping pace with the topics in class? (I have access to the notes and the CAPA's but without feedback from you I can't tell exactly where you are with the material).

2. Are people finding the material both here on the blog and on the main website useful? Any suggestions regarding what is good/bad, could be improved/added/discarded?

Any feedback you have for me would be very much appreciated! Hope your courses are going well, I know this can be a busy time of year!

Question: Current and a switch

I'm just wondering. If there is an open switch AFTER a resistor, the current still goes through the resistor but does not continue in the circuit, right?

If there is an open circuit, there is no current flow. For current to flow there has to be a continuous path from a high potential to a low one, otherwise charge would have to build up somewhere... which cannot happen. If the charge built up in this way, the charges flowing in behind them would feel more and more repulsive force until the potential required to pile in more equaled the potential of the battery and there was no driving force left. This is basically what happens in a capacitor, and a real circuit will have some of this occuring, but the end result in both cases is that the current stops flowing.

Ever been on a really crowded bus? There is somewhat of a driving force from the front of the bus to the back of the bus as people get on and the bus driver yells to everyone to go to the back. As the bus fills up, and fills up, people at the front push people in the middle who push people at the back closer and closer together, but eventually there is nowhere for them to go... at some point, the flow of people towards the back has to stop until people get off and make room again.

A wire is like an already crowded bus. The charges close to the battery (the front of the bus) feel a driving force from the battery, those charges move slightly, and because they are displaced in the wire, there is a net charge which repels like charges near them and pushes them ahead a bit. This continues all around the wire. But, if there is a break in the loop, there is nowhere for the charges to go (they can't just jump out of the wire), so they just sit there and wait until there is a way for them to move.

The short answer to your question is: if there is an open switch anywhere in a loop, there will be no current flowing through any part of that loop. If there are alternate closed paths in the circuit, then current can flow through those parts of the circuit, but they must constitute a closed loop, with some source of potential a.k.a. electromotive force (EMF) (a battery, DC or AC power supply, a charged capacitor...).

Hope this helps. Please post a comment to this if you need clarification

Topic open: Anything DC circuits

The following topic is now open for questions: Anything DC circuits. Including: Resistor networks, Kirchoff's laws, Capacitor networks, RC circuits...

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Using loop rules (Chapter 19 #29)

I am stuck at # 29 (chapter 19) and I really don't know where I went wrong. I made 2 loops, 1 with R1 R2 and another with R2 and R3 and then I had one junction equation.

Well, from what you said it seems like you are on the right track. Using two loop rules, and one junction rule, I get the following equations:

loop 1: E1-I1R1-I2R2=0 loop 2: E2-I2R2-I3R3=0 junction: I1+I3=I2

This does give you 3 equations with 3 unknowns. Solving these systems by substitution can be a bit tricky, but with perserverance can be done. I show an example on the help home solving both by substitution and a matrix method (which can be a quick way to solve if you have access to a graphing calculator, or are good with matrix reduction). To see the example and a review of Kirchoff's laws go to: Kirchoff's laws and their application.

Let me know if these are not the equations you got, and we will backtrack some more and find out where the problem is. (Remember, you can post follow-up questions as comments to this post.)

Question: Adding another resistor

I was wondering if you could tell me if, say you have five resistors in a circuit, and you add another one, does the power of the battery have to increase? Or does the system just slow down or something...

Ok, I'm first going to have to clairfy and simplify the system we are talking about. I'm going to reduce the problem to a battery and a resistor. The only difference in having 5 as opposed to 1 is reducing that network of 5 resistors. We will also assume that we have a battery providing a constant voltage source (this is the normal case), and that it can provide whatever current we demand of it (it has no maximum current output).

So, let's take our battery and resistor. Using Ohm's law, we can determine that the current is I=V/R (if we had 5 resistors of equal resistance in series to start the current would be I=V/5R since the equivalent resistance of those 5 resistors in series would be 5R). And then the power drawn from the battery would be P=IV, or P=V²/R.

Now suppose we add another resistor, of equal resistance, in series. The equivalent resistance of the two resistors would be 2R, and the new current would be I=V/2R, and the new power would be P=V²/2R (if we had added one more in series to our 5 resistors we would have I=V/6R, and P=V²/6R).

The voltage provided by the battery does not change due to the addition of another resistor, however the power output decreases (because the current drawn decreases).

We can go through the same analysis assuming the resistors are in parallel. The current with just one resistor is I=V/R, and power is P=V²/R. If we add a resistor in parallel, the current will become I=2V/R, and the power will be P=2V²/R. Here, again the voltage of the battery does not change (it is a constant source of potential), but the addition of another resistor in parallel increases the current drawn from the battery. Since the current increases, the power supplied by the battery also increases.

So, what happens when we "add" a resistor to a circuit? That depends on how we add the new resistor. The voltage provided by the battery stays constant in all cases. The current may change depending on the arrangement of the resistors. To determine how the current will change you must reduce the resistor network before and after adding in the new resistor, and use Ohm's law to determine the current before and after. If the current changes, the power will change, since Power is proportional to current. Again, compare before and after to determine the change.

Finally, let's be clear about the difference between "voltage" and "power". Voltage, or potential difference, is increase/decrease in potential energy per unit of charge. Power is the energy per unit time, in this case electrical energy per unit time flowing through the circuit. It is the battery which supplies both. That energy has to come from somewhere, right? You can think of power as being an energy transfer, the battery is providing energy at some rate to move charges around the circuit, at some rate.

Hope this helps clarify some things. You might also be intrested in a previous student's question on Voltage, EMF and resistors (about half way down the page). Let me know if this answers your question or not (by posting a comment to this post).

Topic open: Resistor networks

The following topic is now open for questions: Resistor networks.

To pose a question, please post a comment to this post by clicking on the comments link below.

Solution to first homework: a ring of charge

The solution to my first "homework" exercise is posted on the homework page.

Please let me know if you have questions about the solution, or if you think / don't think similar exercises would be helpful in future.

Question: E=V/d or V=-Ed? (to minus or not to minus)

I was wondering why the notes from class say that for uniform Electric fields E=V/d, while the textbook and WebTA say that V=-Ed. Why is there no negative sign in the one case, but the negative is included in the other?

That is a very good question. I can first tell you that the reason my formula and the formula in the book are the same is that I copied mine out of the book. ;) I can also say that Prof. Altounian's notes are correct.

To get to the heart of this matter, really we have to think about what V is and what E is. The electric potential, V, is a scalar quantity. That is, it has a magnitude at a particular point in space, but no direction associated with it. But the electric field, E, is a vector quantity. This means it has both a magnitude and a direction. So if we look at a simple formula for the electric field, like E=V/d, we have to think this isn't the full story... how do we relate a vector and a scalar?? where is the direction part of E??? Well, it isn't there. Really all we get is the magnitude of E. So, whether you stick a negative sign there or not, you have to determine the direction of the electric field in another way (by determing where a high potential is and where a low potential is).

So, why did the text book (and myself) bother putting a negative sign there? Well, it comes from the calculus relation between E and V. If you have a map of V over an area, the electric field points downhill, so when you look at it from within a differential formalism you need the negative sign (so I'm used to seeing it that way).

In any case, I mean to change my webTA pages to agree with the class notes on this point. And the moral of the story is that since E is a vector you have to determine the direction as well as the magnitude.

Server unavailable Jan 27 05:00-07:00

I received this message this morning regarding the physics department servers:

please be advised that tomorrow (friday) the webserver (and its filesystem) will be unavailable between 05:00 and 07:00.

This will effect the webTA home (site with all examples etc), but not the blog site.

Posting questions

Hi all, I would like to suggest that when you ask a question related to a text book problem that you include the problem with your question. It makes it easier for everyone reading your question, and much easier for me if I happen to not be near my text book (or too lazy to look it up right away).

Also, I think everyone is getting the idea about asking questions by posting comments. Don't worry if they don't appear right away, I have to clear them first. I would prefer, unless you are having a problem with the site, that you don't email me with questions. It's just easier to have them all in one place on the blog, and possibly helpful to others that might have been wondering the same thing.

Thanks everyone for your continued participation, so far I think this is working out quite well. :)

Question: 2 Charges Dropped in an Electric Field (Chapter 17, #68)

I am confused on what the equation for conservation of energy should be for this problem....Should we consider potential energy due to the height?

For example I assumed the equation to be: mgh(before)+PE(before)=KE(after)

The question states: "Near the surface of the Earth there is an electric field of about 150V/m which points downward. Two identical balls with mass m=0.540kg are dropped from at height of 2.00m, but one of the balls is positively charged with q₁=650μC, and the second is negatively charged with q₂=-650μC. Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground."

Whoo-ee! That's a nice problem, it incorporates several different ideas, just the kind I hated as a student, and see the value of now when I teach. :)

Your energy equation is mostly correct, but we need to figure out the signs of the energies.

Let's first draw a diagram, because I find that problem a mouth full. (I will put in the electrostatic forces, but not gravitational... we all know that points downward anyway).

So we have the positive charge (drawn on the left) experiencing a downward force due to the electric field, thus as it falls (down) it's electric potential energy will be decreasing, just as it's gravitational potential energy will be decreasing. However, for the negative charge (drawn on the right), the electrostatic force is pointing upward, opposing the falling motion, so it's electric potential will increase as it falls.

We will have to look at the conservation of energy each charge separately. Let's start with the positive charge:

PE(electric)+PE(gravitational)=KE
qEh+mgh=(1/2)mv₁²

(note that I used h for the distance the charge travels through the electric field).

Now let's look at the negative charge:

PE(gravitational)=KE+PE(electric)
mgh=(1/2)mv₂²+qEh

When plugging in the numbers, disregard the sign of the charge. I have included it already by determing whether the electric potential energy increases or decreases. Alternatively you could have used the energy equation for the first charge and plugged in the sign of the charge to get the same thing. Personally, to avoid confusion with signs, I think it's better to work through the logic of what's going on than to plug in signs, but that doesn't make the other way wrong.

So, using the energy equations for each ball, you can solve for the final velocities. I have confidence that you can all plug in numbers, so I will leave it at that. Hope this helps, and if you have any further questions feel free to post a comment.

Question: Increase plate separation, increase energy stored in capacitor

If two plates of a capacitor with constant charge have their separation doubled, the energy stored also doubles. I'm confused as to why this happens, since I thought E-stored was related to C, and C decreased as distance of separation increased. I'm looking at the equation Energy Stored = 0.5CV^2.

You are right that the energy stored in a capacitor if related to C, the capacitance. However, you have to consider what is being held constant as the separation doubles. If you look at the equation: U=(1/2)CV², you will notice that U is proportional to C, however you have a factor of V² that changes in a way you can't determine directly. If instead, you re-write the equation substituting in V=Q/C, so that you have: U=(1/2)Q²/C, then you know that Q is a constant, so we can talk about U being proportional to 1/C. Thus, as C decreases, U increases.

It might help you to look at the example I posted on this if you haven't already: Chapter 17, #50. Note also that since the amount of energy stored in the capacitor is increased, work must be done to increase the separation between the plates.

First homework: a ring of charge

Well, I've posted my first "homework" exercise. I hadn't gotten around to it because I've been trying to polish up some other things, and keep up with your questions. It should be a short exercise, so before you dive into doing a bunch of math, think about ways to simplify the problem.

If you're interested you can get to it from the homework page.

Good luck, I will post an answer in a few days or so.

Topic open: Resistance and Resistivity

Resistance is futile... if less than 1Ω. ~ unknown author

The following topic is now open for questions: Resistance and Resistivity.

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I realize you may still have questions regarding capacitors, feel free to continue asking questions on old topics (by posting comments to the appropriate topic post).

Popups?!?

I would like to know if any of you have been experiencing pop-ups on this page. I have pretty good pop-up blocking, so I had no idea that the "free" counters I use on all my webpages have recently started creating popups on the webpages they are placed on.

If anyone has seen this happen on the blog, or the static tutorial page, please let me know and I will remove the counter immediately.

Thanks, and my appologies to anyone experiencing these popups.

Edit: I have removed the webstats4u counter and replaced it with another that claims it will not put ads on the sites that use it. Still, let me know if anything strange happens. (2:14 PM Jan 23 2006)

Question: Capacitor miscellany

here are my questions about capacitor:

1 #51 b)Why doubling?

The energy stored in a capacitor is given by:

U=(1/2)QV=(1/2)CV²=(1/2)Q²/C.

In Ch. 17, #51 (b) we are told that the amount of charge on each plate is doubled (ie. Q is doubled), and the capacitor remains connected to a battery (ie. V is held constant). So, we can look at an initial energy, U₀, and a new energy after the charge is doubled, U':

U₀=Q₀V
U'=Q'V=(2Q₀)V

The ratio of the new energy to the old is then:

U'/U₀=Q'V/Q₀V=2

So, indeed if the charge on the capacitor is doubled, the energy stored in the capacitor is doubled.

2 Say, we have 2+Q on the left plate and 2 -Q on the rite, can I just add 1 +Q to the left and add nothing to the rite? Or it is a must for me to add the same amount of charges to both sides?

You must add the same amount of charge to both sides. (Check out the Ch. 17 summary on where it describes a capacitor.) If you added more of one charge to one side than the other, the net charge of the capacitor would be non-zero. I'm not sure what exactly would happen, but it wouldn't be a simple capacitor anymore and wouldn't follow the behaviour we've been discussing. Sorry I can't give a better explanation than that.

3 If I added 1 +Q to left and 1 -Q to rite, the total charge should be +2Q, rite?

Well, the net charge would be zero, the total amount of charge would be |2Q| (those || are to show we are talking magnitude/absolute value and disregarding the signs), but the charge on the capacitor (the "Q" you use in the capacitor equation) would be 1Q. The measure of charge that matters in a capacitor is how much you have separated off to each plate, so it is this charge you consider.

4 #77b) what i did was PE/2=1/2*m*v^2, what is wrong with my solution( i know it must be wrong -_-||)?

I had a student ask this question last year, I think the answer given there will help you also. Check out electrostatics: previous student questions, it is the second item on the page.

I believe your problem lies in the factor of 1/2 you are using for the potential energy... I'm not sure where this comes from, and I'm pretty sure it isn't necessary. ;)

Good Luck, hope this helps. As always you can ask follow-up questions.

Topic open: Anything Capacitor

The following topic is now open for questions: Capacitors.

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Question: Back-to-Back Capacitors (Ch 17, #41)

One student asks: "I was able to find the number of charges using Q=CV...but I have no clue how to solve the question."

Another asks: "I figured out the total charge after charging to batteries, but I dunt understand why they made "the positive plates connected to each and the negative plates connected to each other" and I cannot get the answer either. Another question from #41: they said "the potential difference across reach". Does it have the same meaning of the voltage of a battery?"

The question states: "A 2.50-μF capacitor is charged to 857V and a 6.80-μF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potenial difference across each and the charge on each? [Hint: charge is conserved]"

Ok, so it seems there is some degree of confusion about this question and how to handle it. The first step is to calculate the charge on each isolated capacitor using: Q=CV, as suggested by the first student. The total charge when they are put together will be the sum of these two charges. You will have:

Q₁=(2.5e-6)(875)
Q₂=(6.8e-6)(652)
Q_total=Q₁+Q₂

The second question here actually holds a key part of the answer, and again, it is about interpreting the wording of a question. What does it mean when they say "the postive plates are connected to each other and the negative plates are connected to each other"? Let's draw this situation...

Notice that by connecting the terminals of the capacitors with like charge we have essentially placed them in parallel. So, we can treat them like parallel capacitors and find the equivalent capacitance of this arrangement. For capacitors in parallel, we have: C_eq=C₁+C₂.

Now we know the total charge on the two capacitors, and the equivalent capacitance of the two. We can now treat them as one capacitor with the total charge Q_total and equivalent capacitance C_eq to find the new voltage across both (note that the voltage across things in parallel is always equal).

Q_total=C_eqV_new
V_new=Q_total/C_eq

To address the second student's last question regarding the potential difference, the potential difference here is somewhat similar to the voltage across a battery. A battery's job is to raise the potential energy of charges passing through it, it puts more positive charge on one side and more negative charge on the other. The same is happening here, on one side of the two capacitors there is more negative charge and on the other there is more positive charge. So, we say there is a potential difference or a voltage across the two capacitors.

Hope this is helpful, if you have any follow-up questions please post a comment.

Question: One charge, E-field and force

Hi, I was wondering how to calculate both a force and an electric field when you're only given one charge. For example, if I have a positive test charge q, and a positive charage Q1 r distance away from the test charge... I'm confused on how I can calculate the force vector (resultant) in order to set up a triangle and find the x and y components of the force (same for electric fields).

If you are given only one charge, and no electric field anywhere due to some other arrangement of charges, then all you can do is calculate the electric field due to that one charge. In this case, you can determine the electric field at a point some distance (r) away. The direction of the electric field will be the same as a force would be on a positive charge if you put one there, but the magnitude of this test charge (q) is divided out (E=F/q). Remember that for a point charge, the electric field points radially outward/inward from the positive/negative charge. This makes sense if you think about putting a positive test charge anywhere around the charge you are given, as the force between the two will be along the direction of the separation.

You might also be asked about a charge in an electric field (usually a uniform one, since that's easiest to deal with). In this case you can determine the force on the charge due to this external electric field by making use of the definition of electric field: E=F/q. Flip this around, and you have: F=Eq, which can be very useful. Let's say you have an electron in a 1 V/m (N/C) uniform electric field pointing north. The magnitude of the force on the electron will be: F=(1N/C)(1.6e-19C)=1.6e-19N. The direction of the force will be south, since the electron is negatively charged and the direction of E is defined as the direction of force on a positive charge.

I hope this helps, please follow-up if I haven't answered your question here.

sorry!

My appologies to everyone. I did not realize that by turning on a comment moderation feature no comments would appear on the blog until I cleared them. Sorry to everyone who posted a comment up until now! I will try to answer your questions ASAP.

Edit: I will also be notified by email now whenever a comment needs to be cleared, so I shouldn't miss out on your questions again!

Question: Charge transfered between plastic balls (Ch 16, #22)

i have an r, an F and constant k....im not sure how to go about this, but i solve for Q1*Q2, and then not sure where to go from there. if the first ball is uncharged , does that mean Q1 is 0 coulombs?

The question states: "A charge Q is transferred from one ball to an identical ball 12 cm away. The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?"

So, the key thing to pick up on here is that the balls are initially uncharged (Q₁=Q₂=0). This means that after charge is transferred between the two balls the net charge, Q₁+Q₂, will be zero. If we use this fact, we can see that Q₁=-Q₂. So, since a charge is transferred from one ball to the other each now has an equal but opposite charge, let's call it Δq. So, we can say that Q₁ now has a charge of +Δq (a positive charge), and Q₂ now has a charge of -Δq (a negative charge).

Now we can use the equation for the electrostatic force between two point charges to solve for Δq as you suggest.

It is worth noting that there were two ways we could have determined that the balls had opposite charge from what is given in the question. We used the fact that the balls are uncharged, telling us that the net charge after transfer is zero. We could also have used the fact that the force is attractive to determine that the balls are oppositely charged. But... we still would have needed to know that the magnitude of the charge on each was equal from the zero net charge. Often though, the wording of a question contains more information than you might suspect at first glance, so it is worth looking for hints like this.

You might also find the following posted example helpful: Chapter 16, #21 which also involves two charged spheres. The only difference is that in this example the net charge of the two sphere is not zero.

Question: Charged droplet in an Electric Field

we encountered a problem where a charged droplet remains stationary above the earth, the electric field of the earth is given and its asks how many excess electrons are in the droplet.

so i assumed that since there is no movement there has to be no net electric field. so i assumed the electric field of the droplet is also 145 n/c just negative. i used the equation kQ/r^2 = electric field. so i multiplied 145 by the square of the radius and i divided it by the constant k to get the total charge of the droplet. then i divided that by 1.602 E-19 to get the number of electrons present, but i still could not get the right answer, what was i doing wrong?

You are right that there has to be zero net something, however, it isn't the Electric field that cancels to suspend the drop, it is the net Force on the droplet that has to be zero to establish static equilibrium. So, let's think about what forces are exerted on the droplet here. Firstly, there is a force of gravity exerted down on the drop since it is suspended above the Earth. Secondly, because the droplet is charged and is in an Electric field, it will feel a force due to this Electric field. We know that this force must point "up" (ie. oppose gravity), so we can determine from this the sign of the charge. Since the Electric field of the Earth points "down" (radially inward, gravity points radially inward), and the direction of the Electric field is defined as the direction of the force on a positive point charge, the charge of the particle will have to be negative. We can now draw a diagram to make this more clear:

The electric force is F_el=qE. We know E (it is given in the problem), and we are looking for q. For the gravitational force we need to do a bit of work if the mass isn't given directly. Since we know the size (radius) of the drop, we can calculate it's mass from the density, ρ, and volume, V=(4/3)πr³: m=ρV. Then we have all the pieces we need.

It should be noted that the equation for the electric field due to a point charge cannot be used in the way you tried to use it. E=kQ/r² is the Electric field due to a point charge with charge, Q, at a distance, r, away.

Hope this is helpful to you, if you have follow-up questions don't hesitate to comment on my answer here. Thank-you for being the first to ask a question!

Topic open: Electrostatic Potential and Potential Energy

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Topic open: Forces on charges and Electric fields

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