# Calculating Dipole Moments

## Homework Statement

Find the value of the dipole moment of the distribution of charges.

-2q on top of equilateral triangle
+q on each of the other points. Each charge is separated by distance d.

## Homework Equations

p = $$\sum$$ (q_i)(r_i)

## The Attempt at a Solution

p = q(d/2) - q(d/2) +(-2q)(d*$$\sqrt{3}$$/2)

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gabbagabbahey
Homework Helper
Gold Member
Hi LabBioRat, welcome to PF!

Dipole moment is a vector quantity, but it's not clear from your post whether you are treating it as a vector or a scalar, and what coordinate system and origin you are using....

Hi LabBioRat, welcome to PF!

Dipole moment is a vector quantity, but it's not clear from your post whether you are treating it as a vector or a scalar, and what coordinate system and origin you are using....
Gabbagabbahey, thank you for the warm welcome. I am treating dipole moment as a vector quantity in the Cartesian Coordinate system.

gabbagabbahey
Homework Helper
Gold Member
Okay, and where is the origin of this coordinate system?

And, you should make it clear which quantities are vectors by putting them in bolded letters like this: p; and use i, j and k to represent Cartesian unit vectors.

Or, you can use LaTeX: $$\vec{p}$$ , $$\hat{i}$$

Since the total charge of the configuration is 0, the dipole moment does not depend on a specific origin. I choose the origin to be at the (-2q) charge.

gabbagabbahey
Homework Helper
Gold Member
Since the total charge of the configuration is 0, the dipole moment does not depend on a specific origin.
Yes, that's right

However it does still depend on how you choose to orient your coordinate system (although it's magnitude will not) ....if the configuration lies entirely in the xy plane it will have a different dipole moment than if it lies in the xz plane...

I choose the origin to be at the (-2q) charge.
Okay, and when you write everything out in vector notation, what do you get?

(-2q)(0$$\widehat{i}$$ + 0$$\widehat{j}$$)+ q( (-d/2)$$\widehat{i}$$ -d $$\sqrt{3}$$ /2$$\widehat{j}$$ ) +q( (d/2)$$\widehat{i}$$ -d$$\sqrt{3}$$ /2$$\widehat{j}$$ )
= 2q( -d$$\sqrt{3}$$$$\widehat{j}$$ )

(-2q)(0$$\widehat{i}$$ + d$$\sqrt{3}$$ /2$$\widehat{j}$$)+ q( (-d/2)$$\widehat{i}$$ +0$$\widehat{j}$$ ) +q( (d/2)$$\widehat{i}$$ + 0$$\widehat{j}$$ )
= 0q( 0$$\widehat{i}$$ +d$$\sqrt{3}$$/2 $$\widehat{j}$$ )

gabbagabbahey
Homework Helper
Gold Member
Careful, you dropped the -2q in front of the brackets

Doesn't the -2q belong on the outside, just like the other charges?
My origin is now at (0,0), and the position of the (-2q) charge is $$(0, d\sqrt{3}/2)$$.

I do not quite understand your warning.

gabbagabbahey
Homework Helper
Gold Member
You went from -2q(i+$\sqrt{3}$d/2j) to 0q(i+$\sqrt{3}$d/2j)

why did -2q become 0q?

That was after I summed up all of the charges and vectors.
All of the charges evaluated to 0q, and
the vectors from each of the charges all evaluate to ( 0$$\widehat{i}$$ +d$$\sqrt{3}$$/2 $$\widehat{j}$$ )

Thus my final answer would be $$0q( 0\widehat{i}$$ +d$$\sqrt{3}$$/2 $$\widehat{j})$$

I apologize for the confusion.

gabbagabbahey
Homework Helper
Gold Member
That was after I summed up all of the charges and vectors.
All of the charges evaluated to 0q, and
the vectors from each of the charges all evaluate to ( 0$$\widehat{i}$$ +d$$\sqrt{3}$$/2 $$\widehat{j}$$ )

Thus my final answer would be $$0q( 0\widehat{i}$$ +d$$\sqrt{3}$$/2 $$\widehat{j})$$

I apologize for the confusion.
No, you don't add the charges and the vectors separately.... the charges are scalars that multiply each of the vectors:

$$-2q\left(0\hat{i}+\frac{\sqrt{3}}{2}d\hat{j}\right)=-2\sqrt{3}qd\hat{j}$$

$$q\left(\frac{-d}{2}\hat{i}+0\hat{j}\right)=\frac{-qd}{2}\hat{i}$$

$$q\left(\frac{d}{2}\hat{i}+0\hat{j}\right)=\frac{qd}{2}\hat{i}$$

When you add them up, the two vectors pointing in the i direction cancel, but the other one doesn't

Oh my goodness! Thank you for pointing out that fatal mistake!

Help me to sort out this problem:: Prove that, "integration over[J(r)dr]=del(p)/del(t)" ... where p is the electric dipole moment ... please as soon as possible, reply me ...