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Electric field due to point charges and a ring of charge:

  • Thread starter arl146
  • Start date
  • #1
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Homework Statement



A ring of charge with radius R = 0.5 m is centered on the origin in the x-y plane. A positive point charge is located at the following coordinates:
x = 19.0 m
y = -14.6 m
z = -2.6 m

The point charge and the total charge on the ring are the same, Q = +40 C.
Find the net electric field along the z-axis at z = 7.3 m.
E net,x = ?
E net,y = ?
E net z = ?



Homework Equations


Due to ring: E(z)=(kqz)/(z^2+R^2)^(3/2)
Due to a point charge: E= kq/r^2

The Attempt at a Solution



I found the electric field due to the ring pretty easily I just plugged everything into the equation I have above and for that I got 6699.284 N/C, is that right? I thought for the point charge I just plug and chug as well but I got it wrong and I don't know why. I have no idea how else to do it. Help please!
 
Last edited:

Answers and Replies

  • #2
gneill
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20,793
2,773
With charges as large as 40C involved, I doubt that the field strength will be as modest as you've indicated for the ring; I suspect that a few orders of magnitude have gone astray somewhere.

You should be able to "plug and chug" as you you say. Perhaps you could share your calculation.
 
  • #3
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Due to the ring I did:
E(z)= kqz/(z^2+R^2)^(3/2) = (8.988x10^9)*(40x10^(-6))*(7.3)/((7.3^2+0.5^2)^(3/2)) =6699.28362778 N/C.
That's wrong? And for the second part I tried everything, from just plugging in the numbers to using the x and y components of each point. I don't know what else to do.
 
  • #4
gneill
Mentor
20,793
2,773
Due to the ring I did:
E(z)= kqz/(z^2+R^2)^(3/2) = (8.988x10^9)*(40x10^(-6))*(7.3)/((7.3^2+0.5^2)^(3/2)) =6699.28362778 N/C.
That's wrong? And for the second part I tried everything, from just plugging in the numbers to using the x and y components of each point. I don't know what else to do.
So the charge is 40 micro coulombs, not 40 coulombs as you have in the problem statement?

Show your numbers for the point charge calculation. What's the distance vector from the point charge to the location of interest?
 
  • #5
343
1
Yes, it is actually micro coulombs. Sorry!! I called that distance r (little r, not the radius big R) and I did sqrt((19^2)+(7.3^2)) for the E net,.

And for the point charge:
E,x = kq/r^2 = (8.988x10^9)*(40x10^(-6))/(414.29) = 867.7979 N/C
 
Last edited:
  • #6
gneill
Mentor
20,793
2,773
Your r doesn't look right. Once again, what is the vector from the point charge to the test point?
 
  • #7
343
1
Oh ! I actually got it ! Thanks for you time :) I figured it out.
 

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