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.

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

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.

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

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

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

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Topic open: Forces on charges and Electric fields

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

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