Electric dipole question

  • Thread starter hoseA
  • Start date
  • #1
61
0
http://img79.imageshack.us/img79/2859/dipoleelectricfield5nh.png [Broken]

I used the equation E= kqd/(L^3)

L=sqrt(h^2+(d/2)^2)

I got the first answer (for P1).

I can't figure out how to utilize that equation for P2.

The second question asks:

"What is the magnitude of the electric field at
P2? Answer in units of N/C"

what's the value of L (-100... -101.1 or something totally different)... I don't quite understand the variables in the given equation. Help is most appreciated. thx.
 
Last edited by a moderator:

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
4,784
18
The formula you used to calculate P1 is an approximation that is only as precise as the field you calculate is far from the dipole (we say the approximation is valid at large distances). It is useful when the charge distribution we want to calculate is weird (but of charge 0 C in total) and hard to calculate. 100 meters is a reasonably large distance and the use of the approximation would be justified in practice. However, you have all the necessary information at your disposal to evaluate the field exactly.

All you need is Coulomb law.


Note: For P2, the equation you used to approximate P1 is not applicable. In fact, the equation you used is only valid to approximate the field for points on the plane perpendicular to your computer screen and perpendicular to the line joining the charges and such that P1 is a point of that plan. (This defines the plan uniquely) For points in other locations in space, you must use the more general formula

[tex]\vec{E}(r, \theta) = \frac{kqd}{L^3}(2 \cos\theta \ \hat{r} + \sin\theta \ \hat{\theta})[/tex]

where the dipole vector [itex]\vec{p}=q\vec{d}[/itex] is located at the origin and points in the direction of the positive z axis. theta is the polar angle from spherical coordinates that is restricted to [0,pi] (as opposed to [itex]\phi[/itex] that ranges from 0 to 2 pi).
 
Last edited:
  • #3
61
0
quasar987 said:
The formula you used to calculate P1 is an approximation that is only as precise as the field you calculate is far from the dipole (we say the approximation is valid at large distances). It is useful when the charge distribution we want to calculate is weird (but of charge 0 C in total) and hard to calculate. 100 meters is a reasonably large distance and the use of the approximation would be justified in practice. However, you have all the necessary information at your disposal to evaluate the field exactly.

All you need is Coulomb law.


Note: For P2, the equation you used to approximate P1 is not applicable. In fact, the equation you used is only valid to approximate the field for points on the plane perpendicular to your computer screen and perpendicular to the line joining the charges and such that P1 is a point of that plan. (This defines the plan uniquely) For points in other locations in space, you must use the more general formula

[tex]\vec{E}(r, \theta) = \frac{kqd}{L^3}(2 \cos\theta \ \hat{r} + \sin\theta \ \hat{\theta})[/tex]

where the dipole vector [itex]\vec{p}=q\vec{d}[/itex] is located at the origin and points in the direction of the positive z axis. theta is the polar angle from spherical coordinates that is restricted to [0,pi] (as opposed to [itex]\phi[/itex] that ranges from 0 to 2 pi).

Does that mean i substitute pi as the angle (which essentially means 2cospi * sin pi = -2?)

And L= -100^3 and d=1.1 ? correct?
 
  • #4
quasar987
Science Advisor
Homework Helper
Gold Member
4,784
18
hoseA said:
Does that mean i substitute pi as the angle (which essentially means 2cospi * sin pi = -2?)

And L= -100^3 and d=1.1 ? correct?

You mean, supposing you wanted to apply that formula to approximate the field at P2? Then no, the angle to use would be 0. Because since the direction of p defines the positive z axis, P2 is on the positive z axis. But for points along the positive z axis, theta = 0. So the formula boils down to

[tex]\vec{E} = \frac{kd}{L^3}(2\hat{z})[/tex]

And since using this formula implies choosing a coordinate system such that the vector p is resting at the origin, then L would be 100+d/2.
 
  • #5
61
0
quasar987 said:
You mean, supposing you wanted to apply that formula to approximate the field at P2? Then no, the angle to use would be 0. Because since the direction of p defines the positive z axis, P2 is on the positive z axis. But for points along the positive z axis, theta = 0. So the formula boils down to

[tex]\vec{E} = \frac{kd}{L^3}(2\hat{z})[/tex]

And since using this formula implies choosing a coordinate system such that the vector p is resting at the origin, then L would be 100+d/2.
Assuming u forgot the q,
I tried (kqd/L^3)*2 = 7.24951 (this came up as wrong)

I also tried without the q = 1.768174e6.

i substituted (100+(d/2)) for "L".

What am I doing wrong? Thanks for the help.
 
  • #6
61
0
hoseA said:
Assuming u forgot the q,
I tried (kqd/L^3)*2 = 7.24951 (this came up as wrong)

I also tried without the q = 1.768174e6.

i substituted (100+(d/2)) for "L".

What am I doing wrong? Thanks for the help.

any help within the next 10 min. much appreciated. I only have 1 try left. Is there supposed to be a "q" in the equation or is what quasar put correct? thanks.
 
  • #7
quasar987
Science Advisor
Homework Helper
Gold Member
4,784
18
I missed a q yes.
You used q = 1.768174e6? Where does that come from? The text tells you q = 4.1 µC.
 
Last edited:

Related Threads on Electric dipole question

  • Last Post
Replies
1
Views
1K
Replies
6
Views
944
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
16
Views
5K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
3K
Top