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^{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
3. The attempt at a solution
First I broke the electric fields into the x and y axes.
Where, E_{x}=0
For 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.
E+=E=(1/4[tex]\pi[/tex][tex]\epsilon[/tex]_{0})*(Q/(r^{2}+(l^{2}/4)
p=Ql
3. The attempt at a solution
First I broke the electric fields into the x and y axes.
Where, E_{x}=0
For 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.
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