Electrostatics - Point charge question

  • #1
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.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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welcome to pf!

hi joannananaa! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
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 :redface:

there are 3 positive charges, 2 at distance d, and 1 at distance d√2 …

how did you get your figures? :confused:
 
  • #3
Thank you :)

Um...I just plugged them into the formula and wrote one as negative and one as positive. I'm pretty lost.
 
  • #4
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0
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 [tex]7.39 \times 10^6 [/tex] N/C for the two charges along the square edge was right.
However, for the one along the diagonal, your "r" in [tex]E = k \frac{q}{r^2}[/tex] 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.
 

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