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Old 01-28-2007, 01:34 PM   #1 (permalink)
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Physics: Electric Field On A Plane?

Two particles are fixed to an x-y coordinate system:
particle 1 of charge -5.00 microCoulombs lies on the x axis at x = + 6.00 cm and particle 2 of charge +5.00 microCoulombs lies on the y axis at y = + 8.00 cm. Midway between the particles, what is their net electric field, in unit vector notation?

I would know how to do this problem if I had 2 x's but I don't know how this works with the x-y's.
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Old 01-28-2007, 02:13 PM   #2 (permalink)
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just calculate the x and y components separately and add them.
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Old 01-28-2007, 03:04 PM   #3 (permalink)
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so could you say:

1/4piEo x q/r^2 - 1/4piEo x q/r^2
(1/(4pi(8.85x10^-12)) x (-5 x 10^-6)/(.06)^2)i - (1/(4pi(8.85x10^-12)) x (5 x 10^-6)/(.08)^2)j
-1.248e7i + 7024847j
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Old 01-28-2007, 07:32 PM   #4 (permalink)
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Location: Wisconsin
My eyes went crossed when I saw that equation!!
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Old 01-29-2007, 12:06 PM   #5 (permalink)
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anyone want to help out? ive got to get this done in 3 hrs but im a little lost
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Old 01-29-2007, 12:31 PM   #6 (permalink)
Junkie
 
Location: San Francisco
So you need to find the electric field at the point (3.00,4.00). Use superposition to find the net field: add the fields that result from each charge at that point, which you find with Coulomb's Law.
e.g., to find the electric field created by particle 1:
Use Coulomb's Law E = 1/(4*π*ε_0)*(q / r^2) for the x- and y-components, which gives you a vector, (1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2), -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2))

Do the same thing for particle 2 and add the vectors to get your answer. (Do a sanity check on all signs, they can be tricky)

Last edited by n0nsensical; 01-29-2007 at 12:50 PM..
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Old 01-29-2007, 01:01 PM   #7 (permalink)
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so you do (1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2) for particle 1 and give it an 'i' and -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2) for particle 2 and give it a j?

So I dont need to use the -5 microcoulombs? keep it positive?
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Old 01-29-2007, 02:22 PM   #8 (permalink)
Junkie
 
Location: San Francisco
Quote:
Originally Posted by threewingedfury
so you do (1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2) for particle 1 and give it an 'i' and -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2) for particle 2 and give it a j?

So I dont need to use the -5 microcoulombs? keep it positive?
you need to calculate x- and y-components for EACH charge, so in total you'll have 4 uses of Coulomb's Law. The signs are tricky, you do need to keep track of positive and negative charge, but I prefer doing the number crunching in absolute value and applying the signs later. In this case, the field resulting from particle 1 has a positive x-component and a negative y-component because field lines point from positive charge to negative charge. If you draw a diagram with field lines that should be more clear.
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Old 01-29-2007, 08:11 PM   #9 (permalink)
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the wretched forum logged me out and my post was lost, but...

the result I got was 3.6x10^7 N/C and the unit vector was [-.6i+.8j]
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