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Physics Help-GAF

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T

The Lamp

Member
Exuro said:
It's an applet for my masteringphysics course. Here's a link. http://phet.colorado.edu/en/simulation/charges-and-fields

Okay so I'm looking at your work. Shouldn't the y components cancel out? As the force goes away from the positive side it also comes back down, thus they should cancel out no? Maybe I'm just reading it wrong or something.

EDIT: Also found this which looks like nearly the same question. http://www.physicsforums.com/showthread.php?t=334566
Click to expand...

EDIT: Oh wait I think you're right, since the -Q charge would cause an attractive electric field. Oops. I'll check my work for that exact same problem tomorrow but for now I think you're right.
 
E

Exuro

Member
The Lamp said:
EDIT: Oh wait I think you're right, since the -Q charge would cause an attractive electric field. Oops. I'll check my work for that exact same problem tomorrow but for now I think you're right.
Click to expand...
Okay cool, at least I know that. :p So the overall electric field would point left correct? My only concern is at the origin it's pointing to the right while everywhere else it's pointing to the left. After that I'm just not sure how to calculate the magnitude of it. Gonna read over your stuff some more and compare it with whats in my book to try to make something click. It has a few basic integrals, I'm just not sure what would change when its a semicircle instead of a circle and when half of that semicircle is an opposite charge. I appreciate the help.
 
E

Exuro

Member
Okay let me know if this is right or not.

E = kq/r^2

(lambda) = q/rd(theta)

dE = k(lambda)d(theta)/r

dE_x = " "*sin(theta)

Then I integrate from 0 to pi/2 and add that to the integration of pi/2 to pi?
 
L

luoapp

Member
It's equivalent to a positively charged semicircle at II, III quadrants. It's also equivalent to a positively charged infinitely long line placed at (-r, 0), perpendicular to x-axis.
 
T

The Lamp

Member
Exuro said:
Okay cool, at least I know that. :p So the overall electric field would point left correct? My only concern is at the origin it's pointing to the right while everywhere else it's pointing to the left. After that I'm just not sure how to calculate the magnitude of it. Gonna read over your stuff some more and compare it with whats in my book to try to make something click. It has a few basic integrals, I'm just not sure what would change when its a semicircle instead of a circle and when half of that semicircle is an opposite charge. I appreciate the help.
Click to expand...


I'm looking at it now and the process is correct I just had things a bit backwards. The Y components will cancel out, NOT the X. If you draw an attractive force from a -dQ, you can see how the Y components will cancel as you integrate over the semicircle.

Doing physics in Microsoft Paint is annoying so I'll just upload the similar problem I had to do and tell me if you figure it out. If not, I will go into more detail (forgive my scribble handwriting).
In this case, it's the same kind of problem it's just that the semicircle is rotated.

Okay let me know if this is right or not.

E = kq/r^2

(lambda) = q/rd(theta)

dE = k(lambda)d(theta)/r

dE_x = " "*sin(theta)

Then I integrate from 0 to pi/2 and add that to the integration of pi/2 to pi?
Click to expand...

(lambda) is your charge, per unit length.
Your charge is Q, your length is a quarter of a circle.
The circumference of a circle is 2pi*r. Divide that by 4 and you get pi*r/2, so that's your length of a quarter circle.

So lambda = Q/(pi*r/2) = 2Q/(pi*r).

You don't put lambda back into your E or dE by itself. Lambda just represents charge per unit length. You want to replace dQ, which = (charge, per unit length) * (length we're considering for the charge) = (lambda) * (ds) = (2Q/(pi*r)) * (r d(theta))



mPSZd.jpg
 
E

Exuro

Member
Wow that's super helpful. I went to a TA today and I think she taught me it the wrong way. Your drawing makes it pretty clear. So a few question, just to make sure I understand.

The reason you only integrated one quarter was because you knew that the other quarter would be the same magnitude on the point so you could just double it correct?

ds = Rd(theta) In laymans terms, this is a small piece of arc length equal to the radius times a small piece of the angle?

You used sin(theta) just due to where you placed it in the image correct?

Think my only other issues are understanding the relations between Q dQ ect. Only used to deriving/integrating velocity to acceleration so these variables/units are still new to me.

Other than that I think I get it. I kind of copied what you did, but had to flip a few things as this is rotated and mirrored. Mine ended up as positive as it points in the positive x axis. Just want to say thank you. I really appreciate it. Made my day a little bit easier.
 
O

Orayn

Member
I'm really not sure what they want from you for 3.a. either...

3.b. looks like an equipartition theorem problem.
 
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