Electrostatics - Point charge question

1. Three equal positive point charges of magnitude Q = 5.00μ C are located at three corners of a square of edge length d = 7.8 cm. A negative charge -15.00μ C is placed on the fourth corner. At the position of the negative charge, what is the magnitude of the electric field due to the three positive charges?

2. Homework Equations :
F=KQ/R^2

3. The Attempt at a Solution :
I have used the equation K=KQ/R^2 to find that the positive points of charge are 7.4E6 N. I wrote these as <7.4E6, 0> and <0, -7.4E6>. I used this equation on the negative charge as well, and got 11.2E6, which I resolved into X and Y components, which were both 7.9E6 and wrote these as <7.9E6, -7.9E6>.

All of this is just following notes from class, and I am unsure of how to actually get the answer.

Any help would be greatly appreciated.

tiny-tim
Homework Helper
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hi joannananaa! welcome to pf! (try using the X2 icon just above the Reply box )
I have used the equation K=KQ/R^2 to find that the positive points of charge are 7.4E6 N. I wrote these as <7.4E6, 0> and <0, -7.4E6>. I used this equation on the negative charge as well, and got 11.2E6, which I resolved into X and Y components, which were both 7.9E6 and wrote these as <7.9E6, -7.9E6>.

i'm confused there are 3 positive charges, 2 at distance d, and 1 at distance d√2 …

how did you get your figures? Thank you :)

Um...I just plugged them into the formula and wrote one as negative and one as positive. I'm pretty lost.

The presence of the negative charge is irrelevant to the magnitude of electric field due to the other 3 positive charges.
Your E field strength of $$7.39 \times 10^6$$ N/C for the two charges along the square edge was right.
However, for the one along the diagonal, your "r" in $$E = k \frac{q}{r^2}$$ needs to be the diagonal of the square.
Then you just resolve this into x and y components, and add all 3 electric field strengths as vectors to get the E field strength in the 4th corner.