• Support PF! Buy your school textbooks, materials and every day products Here!

Direction of dipole's electric field

  • #1
130
1
Edit: This is probably suited for the lower-level homework section. But I wasn't sure, as this is part of my second year electromagnetism course.

Homework Statement



Two point charges, -q and +q are aligned along the x-axis, such that they are equidistant from the origin and the +q charge is upon the +ve x-axis. Obtain a formula for the electric field at any point from the dipole.

Homework Equations



This is what's given to me in my lecture notes;

E(r)= [3(p [itex]\bullet[/itex] [itex]\widehat{r}[/itex])[itex]\widehat{r}[/itex] - p]/4[itex]\pi[/itex][itex]\epsilon[/itex][itex]r^{3}[/itex]


[itex]\widehat{r}[/itex] is the unit vector pointing in the direction of the point in question, from the origin.

p is the dipole moment, a vector pointing from the negative to the positive charge. It's magnitude is equal to qd. d is the vector pointing from the negative to the positive charge.



The Attempt at a Solution



I can derive this formula fine. What confuses me is the direction/signs of the whole thing.

Let's say I wanted to look at a point P, situated on the positive x-axis a distance r from the origin.

Then we'd have;

[itex]\widehat{r}[/itex] = (1,0,0)

If we sub in this [itex]\widehat{r}[/itex] and the dipole moment qd = (qd,0,0) then we get this overall;

E(r) = (qd/2[itex]\pi[/itex][itex]\epsilon[/itex][itex]r^{3}, 0, 0)[/itex]

So we get an electric field only in the x direction. Makes sense. I don't understand the sign though. This overall electric-field in the x-direction gives a positive result. Essentially saying the electric field is pointing in the positive x-direction, TOWARDS the positive charge. This defies all my logic, the electric field should be pointing in the negative x-direction, surely?
 
Last edited:

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
The expression is only valid for r>d, i.e. outside the dipole. So when you're on the positive x-axis, you're to the right of both charges. Since the positive charge is closer, its electric field will be stronger than the field due to the negative charge, so the net field will point to the right.

P.S. You posted this problem in the correct section.
 
  • #3
130
1
The expression is only valid for r>d, i.e. outside the dipole. So when you're on the positive x-axis, you're to the right of both charges. Since the positive charge is closer, its electric field will be stronger than the field due to the negative charge, so the net field will point to the right.

P.S. You posted this problem in the correct section.
Ah, you're quite right. Deriving the equation uses a Taylor expansion and simplifies everything on the assumption r >> a. Cheers, wouldn't have noticed that.
 

Related Threads for: Direction of dipole's electric field

Replies
4
Views
3K
  • Last Post
Replies
0
Views
6K
  • Last Post
Replies
2
Views
993
  • Last Post
Replies
2
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
8
Views
8K
  • Last Post
Replies
1
Views
1K
Replies
2
Views
7K
Top