# Solve Dipole Problem: Help with E_y = kqd/r^3

In summary, The conversation is discussing how to calculate the electric field for a dipole using the old fashioned method and breaking down the components. They mention the x components cancelling out and trying to simplify the y components. They also discuss the factors of a negative sign and the fact that r>>d. They also mention using Pythagorus to find the length of the hypotenuse and how the first term in the denominator is incorrect.
!dipole!

Okay. So I am trying to do this kind of the old fashioned way. I broke the E's down into x and y component and am adding them together component-wise.

I have found that the x components cancel out. Now I am trying to widdle down the form I have for the y components. As of now I have that:

$$E_y=k[\frac{-|q_1|-|q_2|}{[(d/2)^2+r^2]^{1/2}}*\frac{d/2}{[(d/2)^2+r^2]^{1/2}}$$

The text has it in the form: $$kqd/r^3$$ which I am having a hard time getting.

How do I get rid of the negative sign? AND how do I make use of the fact r>>d??

Anyone??

q1 and q2 are equal.

In your first term you have the denominator to the 1/2. That shouldn't be there. Check your work.

I'm not sure what to say about the negative sign because I'm not sure how you came up with it.

If r>>d then you can ignore your (d/2)^2 term in the denominator because it will be negligible compared to r^2.

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I got the negative sign when I drew the free body diagram of my test charge. Since q_o is +, both fields are in the negative y direction.

Or should I just be considering their magnitude of 2q in this case and forget which direction they point?

hage567 said:
q1 and q2 are equal.

In your first term you have the denominator to the 1/2. That shouldn't be there. Check your work.

It is to the 1/2 because it is the distance from each q to p $c^2=b^2+a^2$
Thanks!

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I got the negative sign when I drew the free body diagram of my test charge. Since q_o is +, both fields are in the negative y direction.

OK, that's just a matter of convention then. You could have made it positive if you wanted.

Or should I just be considering their magnitude of 2q in this case and forget which direction they point?

Since the y components are in the same direction (and equal magnitude), you can just say that E = 2*Ey (where Ey is the y component from one of the charges).

It is to the 1/2 because it is the distance from each
The general equation is $$E = \frac{kq}{x^2}$$ right?

So you found the length of the hypotenuse using Pythagorus, so $$x^2 = (\frac{d}{2})^2 + r^2$$

So there is no need to take the square root to get just x, since you need x^2 in your equation anyway. Does that make sense? Just the first term in your denominator is wrong, the second one is right.

hage567 said:
OK, that's just a matter of convention then. You could have made it positive if you wanted.

Since the y components are in the same direction (and equal magnitude), you can just say that E = 2*Ey (where Ey is the y component from one of the charges).

The general equation is $$E = \frac{kq}{x^2}$$ right?

So you found the length of the hypotenuse using Pythagorus, so $$x^2 = (\frac{d}{2})^2 + r^2$$

So there is no need to take the square root to get just x, since you need x^2 in your equation anyway. Does that make sense? Just the first term in your denominator is wrong, the second one is right.

Oh crap-ass! I do crap like this all the time. . . Thanks hage!

## 1. What is a dipole?

A dipole is a pair of equal and opposite charges separated by a small distance, creating an electric dipole moment.

## 2. How do you solve a dipole problem?

To solve a dipole problem, you need to use the formula E_y = kqd/r^3, where E_y is the electric field, k is the Coulomb constant, q is the charge of the dipole, d is the distance between the charges, and r is the distance from the dipole to the point of interest.

## 3. What is the purpose of solving a dipole problem?

Solving a dipole problem allows us to determine the electric field at a point due to the dipole, which can help us understand the behavior and interactions of electric charges.

## 4. Can the dipole formula be used for other problems?

Yes, the dipole formula can be used for other problems involving point charges or dipoles, as long as the necessary variables are known.

## 5. What are the units for the dipole formula?

The units for the dipole formula are Newtons per Coulomb (N/C) for the electric field, Coulomb's constant (N·m^2/C^2), Coulombs (C) for the charge, and meters (m) for the distances.

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