What Is the Net Electric Field at the Midpoint Between Two Charged Particles?

AI Thread Summary
To find the net electric field at the midpoint between two charged particles, one must calculate the electric field contributions from each charge separately. The electric field from the negative charge at particle 1 should be directed towards it, while the field from the positive charge at particle 2 should point away from it. The distances to the midpoint must be accurately determined, with the correct use of unit vector notation. The calculations involve using Coulomb's law and adjusting for the distances, which are half the distances from each charge to the midpoint. Clarifications on the unit vector directions and distance measurements are essential for accurate results.
threewingedfury
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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?

So I got:
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

Is this anywhere near right?
 
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Not quite.

The electric field vector must point along the line from the positive charge to the negative charge, so one needs to determine the unit vector along that line.

The problem asks for the field at the midpoint between the charges, and so one needs the distance between the positive charge and the midpoint, and the negative charge and the midpoint. For one point at (x, 0) and the other at (0, y), think of (x/2, y/2).
 
so can anyone give me a better idea?
 
Last edited:
pretty please?
im a bit confused
 
So is it more like:

(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?

Since that's r/2?
 
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