# Finding the Net Electric Force on a Charge

## Homework Statement

Three positive point charges are arranged in a triangular pattern in a plane, as shown below.
The Coulomb constant is $$8.98755*10^9 N · m^2/C^2.$$

http://img150.imageshack.us/img150/1052/electricforceproblemqp7.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

Find the magnitude of the net electric force on the 6 nC charge. Answer in units of N.

## Homework Equations

$$F = \frac{K(qQ)} {r^2}}$$

## The Attempt at a Solution

Pythagorean Theorem: $$\sqrt{(6)^2 + (6)^2} = 8.485$$

$$\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(2*10^{-9}C)} {(8.485m)^2}} = 1.498*10^{-9} N$$

$$\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(9*10^{-9}C)} {(8.485m)^2}} = 6.7411*10^{-9} N$$

$$sin(45)(1.498*10^{-9} N + 6.7411*10^{-9} N) = 5.8259*10^{-9} N$$

Unfortunately that's not the right answer

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## Answers and Replies

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You made a good start so far. The magnitude of the force on the 6 nC particle due to the other two particles is done correctly. Your next step is wrong.

Do the following: Draw the two force vectors on the 6 nC charge, and then determine their x- and y- components. Now add the components together to get the net force components on the 6 nC charge. Apply the Pythagorean theorem to these components to get the magnitude of the net force.

Attempt #142,032:
$$F1 = cos(45)1.498*10^{-9} N$$ x $$+ sin(45)1.498*10^{-9} N$$ y
$$F2 = -cos(45)6.741*10^{-9} N$$ x $$+ sin(45)6.741*10^{-9} N$$ y
$$Fnet = 3.706*10^{-9} N$$ x $$+ 5.825*10^{-9} N$$ y

$$||Fnet|| = 6.905*10^{-9} N$$

Thanks Guys!

One of the y components should be negative, not one of the x components. The magnitude still comes out the same.