- #1

n387g

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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/r

E+=E-=(1/4[tex]\pi[/tex][tex]\epsilon[/tex]

p=Ql

First I broke the electric fields into the x and y axes.

Where, E

For the E

Then, substituted p for (Q*l) and then r

I assume there must be more to it than that, but I'm at a loss.

^{4}. Measure r from the -2Q charge and assume r>>l.E+=E-=(1/4[tex]\pi[/tex][tex]\epsilon[/tex]

_{0})*(Q/(r^{2}+(l^{2}/4)p=Ql

## The Attempt at a Solution

First I broke the electric fields into the x and y axes.

Where, E

_{x}=0For the E

_{y}, I used the equation E=(1/4[tex]\pi[/tex][tex]\epsilon[/tex]_{0})*(p/r^{3})Then, substituted p for (Q*l) and then r

^{3}for r^{4}I assume there must be more to it than that, but I'm at a loss.