# Finding the electric field of a electric quadrupole

n387g
One type of electric quadrupole consists of two dipoles places end to end with their negative charges (say) overlapping; that is, in the center is -2Q flanked (on a line) by a +Q to either side. Determine to electric field, E, at points along the perpendicular bisector and show that E decreases as 1/r4. Measure r from the -2Q charge and assume r>>l.

E+=E-=(1/4$$\pi$$$$\epsilon$$0)*(Q/(r2+(l2/4)
p=Ql

## The Attempt at a Solution

First I broke the electric fields into the x and y axes.
Where, Ex=0
For the Ey, I used the equation E=(1/4$$\pi$$$$\epsilon$$0)*(p/r3)
Then, substituted p for (Q*l) and then r3 for r4
I assume there must be more to it than that, but I'm at a loss.

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## Answers and Replies

cupid.callin
E+ is not = E-

n387g
Okay, so you have to add the electric fields at the +Q, -2Q and +Q.
So for +Q, E=k(Ql/(l2+r2)^(1/2)) (cos(theta)rx+sin(theta)ry)
And the for -2Q, E=k(2Ql/r2)(sin(theta)ry)
And for the other +Q, E=k(Ql/(l2+r2)^(1/2)) (-cos(theta)rx+sin(theta)ry)
Is this the correct way to do this?

cupid.callin
+Q, E=k(Ql/(l2+r2)^(1/2)) (cos(theta)rx+sin(theta)ry)

Whats that l after Q

n387g
it's a lowercase L, for length

n387g
Using the equation where p=Ql

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