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Integration of an arc of charge

  • Thread starter vysero
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Homework Statement



Two arcs of charge are center at the origin. The arc at radius r has a linear charge density of +(lambda) while the arc of radius 2r has a linear charge density of -(lambda). (r = 5cm, lambda = 1nC/m, theta = 40°)

a) Calculate the magnitude and direction (as an angle from the x axis) of the electric field at the origin.
b) Calculate the electric potential at the origin.
c) Calculate the work done to bring +1 nC of charge from infinity to the origin.
d) Calculate the magnitude and direction of the electric force on +1 nC of charge when placed at the origin.

problem.png

[/B]

Homework Equations



For A (I think) : Ey = 2kq/(pi)r^2
this is after I integrated from 0 to pi/2 with respect to theta[/B]

The Attempt at a Solution



Well the problem I am having is with this lambda. I calculated E(y) to be:
2kQ/pi(r)^2 where Q = 1x10^-9 and r = .05 m
With this I came up with 7192 n/c for the little arc and -719200 n/c for the larger arc. I then made the assumption that E(tot) = the addition of the smaller and larger arc which is:
726392 n/c @ 90° from the x axis

However I get the feeling that I cannot say Q = 1nC because its 1nC per m... I am not sure how to deal with lambda. I have only approached a so far because I am confused by lambda. I found similar problems and I could do this problem if I knew the total charge of the rod. Well I could do it for each arc individually, however the two combined has me a bit confused. I am not sure if I should add the respective E fields together or square them and take the square root of the product... any help would be appreciated.[/B]
 

Answers and Replies

  • #2
DEvens
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Linear charge density of X per meter means: If you had 1 meter of arc you would have X charge. If you had 0.5 meter of arc you would have 0.5 X charge, and so on.

To get the electric field you have to remember that it is a vector, and you need to integrate. But you said you integrated 0 to pi/2. Who says those are the limits? That is not what is in the diagram.

Symmetry will tell you some nice things. In particular, there is one part of the vector you do not need to calculate.
 
  • #3
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Linear charge density of X per meter means: If you had 1 meter of arc you would have X charge. If you had 0.5 meter of arc you would have 0.5 X charge, and so on.

To get the electric field you have to remember that it is a vector, and you need to integrate. But you said you integrated 0 to pi/2. Who says those are the limits? That is not what is in the diagram.

Symmetry will tell you some nice things. In particular, there is one part of the vector you do not need to calculate.
Right I don't need to calculate the x components of the vectors because they cancel. So I guess I am wondering what the bounds on my integral should be and do I need to figure out how long the rod is in cm or?
 
  • #4
BvU
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You already figured out the x components cancel, but it's even nicer if you show it too. Isn't more work, because you need to set up some integration anyway to establish the y component. And the integration is over theta. Make a little drawing where you calculate the contribution to Ey at (0,0) from a small section of arc (radius r) between ##\theta## and ##\theta + d\theta##. It can be treated as a point charge with charge ##\lambda d\theta## . That should also make clear from where to where ##\theta## should run. As DE says, you can benefit from symmetry if you pick ##\theta = 0## in a smart way.

And the ratio between your radii is 2, not 10 !
 

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