What is the electric field at a point midway between two point charges?

AI Thread Summary
The discussion centers on calculating the electric field at a midpoint between a -8.50μC and a +6.38μC charge, which are 7.79 cm apart. The net electric field is derived from the individual fields produced by each charge, using the formula E = (k*q)/r^2. The calculations initially yield a net electric field of -1.257x10^7 N/C, which is not among the answer choices. The confusion arises from the treatment of the negative charge's field direction, leading to the realization that the magnitudes of the fields should be added due to their directional contributions. Ultimately, the correct approach confirms that the electric fields from both charges combine to yield a total of 8.82x10^7 N/C.
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Homework Statement


What is the electric field at a point midway between a -8.50μC and a +6.38μC charge 7.79 cm apart? Take the direction towards the positive charge to be positive. Assume no other charges are nearby.


Homework Equations


E(net) = E(1) + E(2)

E = (k*q)/r^2


The Attempt at a Solution


E(1) = (8.99 x10^-6 * 6.38x10^-6)/ (0.03895)^2
=3.78x10^7

E(2) = (8.99x10^-6 * -8.5x10^-6)/ (0.03895)^2
= -5.037x10^7

E(net) = 3.78x10^7 - 5.037x10^7
=-1.257x10^7 [N/C]

This is not one of the answer choices, and I am not sure what I'm doing wrong. I thought I understood how to do the problem, so it might just be a complete misunderstanding of what the question is asking? I'm not sure...
 
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Oh, I added the two final numbers together:

3.78x10^7 + 5.037x10^7 = 8.82x10^7

and got the right answer. But I don't understand - why would you add the negative charge?
 
A negative charge on the left exerts a field from right to left. A positive charge on the right does likewise, so the field strengths add.
 
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