"Potential" for confusion: Electrostatic potential and potential energy

Let's start at the beginning:

Electric Potential Energy, like any other potential energy, is defined in relation to a conservative force: in this case, the electrostatic force. If one moves a charge from one point to another through an electric field, one must exert a force over this distance... hence work has been done, which corresponds to a difference in the potential energy.

The electric PE at a given point is generally defined relative to a point at infinity; ie. infinitely far away from the influence of any other charges. This is why we talk about "bringing charges in from infinity" to calculate the potential energy of an arrangement of charges.

One additional complication with electric potential energy compared to graviational potential energy is that charges have different signs. That means that the forces which lead to the potential energy can be either attractive or repulsive. It is perhaps worth drawing from the vector definintion of work:



which tells us that we consider only the contributions of the force which are parallel (or antiparallel) to the path. The result? If we are moving the particle on a path parallel to the force on that particle (the force is acting in the same direction we are moving the particle), the work done by the force will be positive, and the potential energy will decrease. Since we are starting at infinity where the PE is zero, this means we will end up with a negative PE. Conversely, if we are moving the particle in countering the force (the force and direction are anti-parallel) then the resulting PE will be positive.

In this way we can think of PE around charges as hills and valleys: if the force between the "active" particle and another in the arrangement is attractive we will have a potential energy valley, but if the force is repulsive we have a potential energy hill. Let's hold onto this landscape idea and revisit it in relation to electric potential.

Electric, or electrostatic, Potential is the electrostatic potential per unit charge. Think of it as: electric potiential is to electrostatic PE, as electric field is to electrostatic force. That means that the electric potential takes on all the same characteristics as the electrostatic potential: it is a scalar quantity, it can be positive or negative depending on whether the interaction is repulsive or attractive.

Like the electric field, the sign will be determined by considering a positive test charge. With electric field, the direction of the vector quantity is determined by the direction of a force on a positive test charge. Since electric potential doesn't have a direction, it is just the sign which is determined by the positive test charge.

Let's return to our landscape idea then... with PE, we have to consider the magnitude and sign of the charge we are describing, however, since electric potential is per unit charge, it will always remain the same (unless the charges defining the landscape move). Since the positive test charge will be attracted by negative charges, there will be "valleys", or regions of negative potential near these, and near positive charges there will be "hills", or regions of positive potential. In this way, electric potential is kind of a measure of the attractiveness and repulsiveness of a position... just remember that it will be opposite for a negative charge.

I hope I've helped, and not muddled the situation further. Please post a comment if you wish some clarification.

I have some other resources posted for you on the topic of electric potential and potential energy for further reading:

• Descriptions of force, electric field, potential energy and potential
  
• Previous student question: Roadblock on potential and potential energy
  
• Question: E=V/d or V=-Ed? (to minus or not to minus)
  

Topic open: Electric Potential and Potential Energy

The following topic is now open for questions: Electrostatic Potential and Potential Energy.

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

Vector Addition

Since many of you seem a little uncertain about adding vectors, here's a quick review example:



The proceedure is the same when you have more than two vectors. Break down all the vectors into components relative to some coordinate system (note that you can choose this such that it is aligned with one of your vectors), add all the components and then using the pythagorean theorem find the magnitude of the resultant vector, and find the angle using trigonometry.

Topic open: Forces on charges and Electric fields

The following topic is now open for discussion: Forces on charges and Electric fields.

If you have a question regarding this topic, please post a comment to this post by clicking on the comment link below.

Welcome to the physics 102 webTA blog: winter '07

This will be the second year of the webTA, so I hope that things will continue to improve and be more helpful for you all. There will be 3 parts of the "webTA" persona:

1. The static site
   
2. The blog (right here)
   
3. The WebCT discussion forum
   

The static site will stay mostly the same as last year. I hope to add maybe one or two new examples written out in full, however, most new material will stay here on the blog.

The blog will be continually updated with answers to your questions. Old posts will never be taken down so that you can learn from the puzzlings of your predecessors. As I've upgraded to the newer version of blogger, there are now labels for all the topics so things should be easier to find. You can also use the blog search to in the toolbar at the top to search for more specific things within the blog site. I will periodically open the floor for questions via a "topic opening" post to which you can add a comment with your question on that topic. Until such time as it becomes a problem (and I hope it won't), you may post anonymously. I would most prefer you ask questions this way as this allows other students to see what others are finding difficult. Also please read these guidelines for posting questions.

I will also hopefully have access this year to the webCT discussions. This would allow me to pop in periodically and sort of "chat" with you and give me a better idea of where you need help, as well as allow me to give quick explanations to little things you have questions about. I won't be a moderator (I don't have that kind of time), but if I see inappropriate discussions I will have to mention it, so please moderate your selves and your peers. If I notice that lots of people are having the same problems I will likely write up a blog post here and direct you all to it. If I do get access to the discussions, I may set up some sort of "virtual office hours" where I will regularly log in, and try to make them outside of regular tutorial or office hours (say, for example, a weekend time).

You can also contact me by email (in the sidebar). I have a few requests however... first, please use a mcgill address for this (I want to know emails are from students in the class and not random people who found the site), second, it is quite unlikely I will be able to reply to each email individually. If you have a question, then it may be answered on the blog, and I will let you know when an answer is posted. If you have a technical question I will try to make sure it is resolved, and may ask you to try again.

A note on questions: not all will be answered. This depends highly on the volume of traffic and questions received, however, last year (especially at the end of term... hint, hint... don't wait!) I could not keep up entirely. Please be patient, as I might not get to them immediately, and I will give priority to groups of similar questions by many students. The "webTA" is one person (me) not a group of TA's, and as such there is a limit to the amount I can do and how fast I can do it. ;)

One last thing... I'd like the webTA sites to act as an additional resource for the class. Please make use of your tutorials; there's nothing like talking to a real person when you need help. With that, I hope you find the webTA project helpful, and I'm always listening for new suggestions!

Good luck!

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