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Champion_Munch
2005-Oct-01, 02:41 AM
I'm studying for a physics test this week, and I've come across a question that I can't work out.

A part of this particular exam is about Electrostatics, and the question relates to the field intensity in a uniform electric field:

"A small sphere of mass 0.005kg and charge +3 C is floating between two horizontal parallel plates 30m apart. What must be the potential difference between the plates?"

I've looked through my notes, the text book and some sheets and I can't find an equation that suits this question. Any ideas?

EDIT: Whoops, just noticed I posted this in the wrong section...can someone move it please?

with regards

Grey
2005-Oct-01, 03:20 AM
"A small sphere of mass 0.005kg and charge +3 C is floating between two horizontal parallel plates 30m apart. What must be the potential difference between the plates?"I think that the problem might have been been worded, but let me guess what they mean here. Presumably they mean that this is happening on Earth, and they've given the mass of the sphere, so you can work out the gravitational force pulling it downward. Since they describe it as "floating", they presumably mean that this gravitational force is exactly balanced by an electric force directed upward. So you know what the electric force is on a particle with a known charge, which allows you to determine the electric field in the region between the plates (more assumptions that I would have stated explicitly is that the plates are conductors and each is held at a constant potential, and that the sphere has a small enough charge that it isn't significantly affecting the field). Then, knowing the separation between the plates, you should be able to work out what potential difference would create such a field.

publius
2005-Oct-01, 03:26 AM
If the little sphere is floating, then forces must be balanced, of course. It's weight is mg. So there must be an upwards electric force equal to its weight. Since the sphere is small, consider it a point charge.

Now, find what value of E is required. Now, from the expression you have for the field of parallel plates, find the voltage drop across them necessary to produce this field.

Just to rain on this elementary electrostatics parade :), I must point out that a charge of 3C is *huge*. In the "real world", the presence of such a charge, even on a small sphere would perturb the field of two parallel plates held at constant voltage significantly (and hence alter the surface charge distribution), and so the forces would be very different from the ideal case being presented of a point charge in a uniform field. But we could say we are holding the surface charge distribution fixed, and then the above answer would be correct (for a sphere small relative to the distance between plates).

While Coulomb's law is very simple, in the "real world", electromagnetic forces get quite complicated to calculate.

-Richard

Wolverine
2005-Oct-01, 03:36 AM
EDIT: Whoops, just noticed I posted this in the wrong section...can someone move it please?

Voila.

Champion_Munch
2005-Oct-01, 03:50 AM
Thanks for the replies. :) The answer to the question is 0.49V, but I still don't understand how to get to it (sorry, but I'm a big newbie to this stuff).

with regards

Kaptain K
2005-Oct-01, 04:01 AM
"I must point out that a charge of 3C is *huge*

Nowhere does it say 3C, just +3. I assumed (and the answer - given in a later post - seems to confirm) that sphere had a net deficit of 3 electrons.

Champion_Munch
2005-Oct-01, 04:21 AM
Just another question I was doing and ended up with a completely wrong answer...

Q15.
An electron in an electron gun is accelerated from rest in a uniform field by a potential difference of 500 V over a distance of 25cm. Find the speed of the electron as it leaves the elctron gun by each of the following methods:
1 - find electric field strength, hence the force on the electron, hence the acceleration, and hence speed of the electron
2 - find the work done by the field on the electron, hence what kind of energy will it have, use the equation for this kind of energy to find the speed.

I have attempted method 1 so far, and ended up with an answer that was way off-target. This is what I did:

E = ΔV/d
= 0.25/500
=5x10^-4

E = F/q
F = E x q
= 5x10^-4 x 1.6x10^-19
= 8x10^-23 N

F = ma
a = F/m
= 8x10^-23/9.11x10^-31
= 8.78x10^7 m/s^2

a = m/s^2
s = √m/a
= √0.025/8.98x10^7
= 166.85 seconds

s = d/t
= 0.025/166.85
= 1.5x10^-4 m/s

But the answer is 1.33x10^7 m/s :(

Thanks for any help. :)

with regards

publius
2005-Oct-01, 04:47 AM
First thing I see is you got the E expression upside down. 500V over 25 cm is 500/.25 = 2000V/m. Plug that in to a = F/m = q/m * E.

Now, for uniform acceleration, velocity, v = a t, distance, x = 1/2 a t^2. Solving for t,

t = sqrt(2x/a) (Sorry, if it's possible to use pretty math fonts here, I don't know how). Plug that into the velocity expression, and you get velocity as function of distance travelled:

v = sqrt(2xa). Checking units, distance times acceleration is velocity squared.

Plug that in and you should get the correct answer.

Doing it the energy way, you know the expression for kinetic energy. The work done by the field is simply qV, where V is the 500V drop that accelerated the electron. Equate the two.

-Richard

Champion_Munch
2005-Oct-01, 04:59 AM
Argh! Thanks alot...I hate it when it's something really really obvious but you can't see what it is no matter how long you look at it... thanks. :)

To get random symbols, click on Start, Run and type in charmap.

with regards

publius
2005-Oct-01, 04:59 AM
Kaptain K,

I may be hallucinating, but I thought he stated a charge of "+3 C". We would need a large charge to balance a weight of 5 grams to keep the voltage reasonable.


-Richard

Champion_Munch
2005-Oct-01, 05:25 AM
One thing I don't get that you mentioned was -


x = 1/2 a t^2. Solving for t,


What's 'x'? The only formula I remember doing that was similar to that was s = ut + 1/2 at^2.

Scroll up in case you missed my previous response. And I don't think you're hallucinating. ;)

with regard

crosscountry
2005-Oct-01, 05:28 AM
cool, can I post my graduate electrostatic problems on here? what about when things start moving?


man, i'm going to ab-use this sight. YEA

Champion_Munch
2005-Oct-01, 05:31 AM
cool, can I post my graduate electrostatic problems on here? what about when things start moving?


man, i'm going to ab-use this sight. YEA

Sorry, I'm just having a little trouble with a few questions....and it's only grade 11. :)

with regards

crosscountry
2005-Oct-01, 05:37 AM
no problem, it's just that I hadn't considered this a resource for those kind of problems.



next time I won't hesitate.

Champion_Munch
2005-Oct-01, 05:46 AM
I wouldn't usually do this, but I don't have the phone number of anyone in my physics class, and I won't have much time after the weekend to spend on this. :(

with regards

Grey
2005-Oct-01, 01:30 PM
cool, can I post my graduate electrostatic problems on here? what about when things start moving?I don't think there's any problem with using this site as such a resource. Heck, in graduate school, the problems are rough, and we were generally encouraged to work together on them. I think you should be clear that it's a homework problem, and, as seen here, most people will respond by trying to point you in the right direction without just giving you the solution, just as a good tutor should.

crosscountry
2005-Oct-01, 02:18 PM
I can appreciate that. some of these problems are very difficult.




anyone here good at Statistical Mechanics? I sure am NOT.

Eroica
2005-Oct-01, 03:11 PM
"A small sphere of mass 0.005kg and charge +3 C is floating between two horizontal parallel plates 30m apart. What must be the potential difference between the plates?"
1: The weight is mg = 0.005 kg * 9.84 ms-2 = 0.0492 N

2: The Electric force is qE = 0.0492 N (q = 3 Coulombs - sorry, Kaptain K!)
=> E = 0.0492/3 = 0.0164 Vm-1

3: Now, V = Ed (http://scienceworld.wolfram.com/physics/FiniteParallelPlanes.html) for finite parallel planes, where d is the separation (d = 30 m)
=> V = 0.0164 * 30 = 0.492 Volts

which is the answer you were given (rounded to two decimal places).

QED

Edit: typo

publius
2005-Oct-01, 04:11 PM
Crosscountry,

Feel free to post graduate level electrostatics. I won't be much help, but feel free. :)

You know, when I was a young buck, I used to hear who at the time I would've called "old coots" talking about how "they used to be good" and all the stuff they used to know but have forgotten. I thought that was a bunch of crap.

Well, guess what? I'm in the same place now myself. :lol: I've got a copy of Jackson (the "bible" of classical electrodynamics), one by Swinger, and about 10 others. I used to know it and used to be good. Now.....<sigh>


-Richard

publius
2005-Oct-01, 04:45 PM
Champion,

Nearly forgot your question about what 'x' was.

'x' is just the variable I used for distance, so the function x(t) is the particle's position as a function of time (under uniform acceleration). In 3-space, for cartesian coordinates it is common practice to use 'x, y, z' as the 3 spatial variables, and for one dimension, 'x' is the first one to come to mind.

This is the same as other formulas you've seen. I did not include an initial velocity (or position) in that. Say this is 11th grade? I don't know if you've studied Calculus yet or not, but when you do, deriving the formulas for uniform acceleration will be a piece of cake.

-Richard

crosscountry
2005-Oct-01, 05:24 PM
Crosscountry,

Feel free to post graduate level electrostatics. I won't be much help, but feel free. :)

You know, when I was a young buck, I used to hear who at the time I would've called "old coots" talking about how "they used to be good" and all the stuff they used to know but have forgotten. I thought that was a bunch of crap.

Well, guess what? I'm in the same place now myself. :lol: I've got a copy of Jackson (the "bible" of classical electrodynamics), one by Swinger, and about 10 others. I used to know it and used to be good. Now.....<sigh>


-Richard


I find Jackson's style a little dense. Griffiths does really well for undergraduate. I use that to weed through Jackson's.

genebujold
2005-Oct-01, 08:52 PM
An electron in an electron gun is accelerated from rest in a uniform field by a potential difference of 500 V over a distance of 25cm...

Well, personally, I find that barrel to be far too short to get any real accuracy, and second, I prefer black powder to all them 'lectrons and "potentials" and what not...

Champion_Munch
2005-Oct-02, 01:13 AM
Thanks alot guys. :D

with regards

Tobin Dax
2005-Oct-02, 06:45 AM
anyone here good at Statistical Mechanics? I sure am NOT.

Nobody's good at Stat Mech. :)

crosscountry
2005-Oct-02, 04:25 PM
Boltzman couldn't even get it right.

adiffer
2005-Oct-03, 09:05 PM
I remember pounding my head with Jackson's book. Later on he showed things from a 4-D perspective and the lightbulb went on. I sure wish he had written it the other way around.

StatMech is just plain weird.

crosscountry
2005-Oct-03, 11:24 PM
wow, I agree.

Grey
2005-Oct-04, 01:24 AM
I remember pounding my head with Jackson's book. Later on he showed things from a 4-D perspective and the lightbulb went on. I sure wish he had written it the other way around.Actually, I think it's just that a certain amount of pounding is required. Perhaps after sufficiently many impacts, the concepts are finally jarred loose from the text, and fall out into your brain. :)

Champion_Munch
2005-Oct-04, 05:38 AM
He's decided to put the test forward another week (can't remember the reason) so now I have until next wednesday. :D

with regards

adiffer
2005-Oct-04, 05:50 AM
I had a professor in grad school who came down to my office during my first year and told me I wasn't going to make it. He explained that the best put in more than 70 hours a week and it was only in the stuff past 70 that they got anyting useful done. He didn't think I was willing to put that much effort in. Of course, I disagreed, but I also silently vowed to start tracking my time.

During my dissertation research, I keep a log/journal with dates and times. It turned out that at my peak, I came real close to his prediction. I would have done it except for diversions to deal with my chronically underfunded bank account. I also understood that I needed many, large blocks of uninterrupted hours to do the work. In the end I agreed with him and told him so. It was several years later and he had forgotten all about it. 8)

I no longer think that pounding one's head is good. My best work involved concentration, but no significant frustration. Annoyance at some bit of work is typical, but frustration shows that you have to change things to get back on focus.

I look back at that time fondly, but there is no way I'd willingly go through that hell again. 8)

crosscountry
2005-Oct-04, 01:47 PM
that's why grad students are often younger. they don't know better.


I'm in that category.