- #1
Saladsamurai
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[SOLVED] Electric Field due to a Dipole
The figure shows two charged particles on an x axis: -q=[itex]-3.20(10^{-19})C[/itex] and q=q=[itex]3.20(10^{-19})C[/itex]. They are both a horizontal distance of 3 m from the y axis. What are the magnitude and direction of the net electric field produced at P at y=4 m?
Okay. So I know I can use the Dipole equation for this, but I had originally tried placing a test charge at P and adding the fields component wise.
I got the wrong answer and I was just wondering why you cannot take this approach?
This is what I had tried:
[tex]E_{px}=\sum E_x=k[\frac{-|q_1|-|q_2|}{d^2}](\frac{3}{5})=-3.836(10^{-10})[/tex]
And
[tex]E_{py}=\sum E_y=k[\frac{-|q_1|+|q_2|}{d^2}](\frac{4}{5})=0[/tex]
Homework Statement
The figure shows two charged particles on an x axis: -q=[itex]-3.20(10^{-19})C[/itex] and q=q=[itex]3.20(10^{-19})C[/itex]. They are both a horizontal distance of 3 m from the y axis. What are the magnitude and direction of the net electric field produced at P at y=4 m?
Okay. So I know I can use the Dipole equation for this, but I had originally tried placing a test charge at P and adding the fields component wise.
I got the wrong answer and I was just wondering why you cannot take this approach?
This is what I had tried:
[tex]E_{px}=\sum E_x=k[\frac{-|q_1|-|q_2|}{d^2}](\frac{3}{5})=-3.836(10^{-10})[/tex]
And
[tex]E_{py}=\sum E_y=k[\frac{-|q_1|+|q_2|}{d^2}](\frac{4}{5})=0[/tex]