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Silversonic

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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.

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.

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

E(

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;

If we sub in this

E(

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?

## 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 q**d**.**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 q**d**= (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?

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