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

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SUMMARY

The electric field at a point midway between a -8.50μC charge and a +6.38μC charge, separated by 7.79 cm, can be calculated using the formula E(net) = E(1) + E(2). The calculations yield E(1) = 3.78 x 10^7 N/C and E(2) = -5.037 x 10^7 N/C, resulting in E(net) = -1.257 x 10^7 N/C. The confusion arises from the addition of electric fields due to opposite charges, where the direction of the electric field from the negative charge is towards itself, thus contributing negatively to the net field.

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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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