HueyFreeman
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Homework Statement
There is an uneven arrangement of electrons and protons on a circular arc of radius r = 2 cm with an electron (E1) at 0 degrees, a proton (E2) at 30 degrees, an electron (E3) at 80 degrees, an electron (E4) at 130 degrees, and a proton (E5) at 160 degrees.
What are the direction and magnitude of the net electric field.
Homework Equations
E = E1 + E2 + E3 + ...
E = k|q|/r^2 r-hat
The Attempt at a Solution
I calculated the amount of force a single particle exerts.
E1 = k|q|/r^2 = (8.99E9*|1*1.6E-19|)/(0.02^2) = 3.596E-6N
All particles should exert the same magnitude of force, correct?
Then I split the force magnitudes into their vector components.
For the proton E2 this looks like E2 = (3.596E-6cos(30))i + (3.596E-6sin(30))j, and I repeated this for all the charges
I added all these forces vectorially - and I got a total of 1.644E-6i + 9.324E-6j
After this step I use the pythagorean a^2 + b^2 = c^2 identity to figure out the magnitude length of the force vector.
This gives me an answer in N, but I want an answer in N/C, so I divide by (5*1.6E-19), because I have 5 particles and are the same distance from the point charge.
It's wrong, though, by a gigantic order of magnitude.
I don't know what to do. We are given an answer that says 160N/C towards P (which is not specific anyways, since there are 3 protons).
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