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fallen186

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## Homework Statement

-----------------------------------------

A charge of -1.0 µC is located at the origin, a second charge of 2.6 µC is located at x = 0, y = 0.1 m, and a third charge of 13 µC is located at x = 0.2 m, y = 0. Find the forces that act on each of the three charges.

I need to find the y forces on the charge 2.6 µC. I have one last try.

## Homework Equations

-----------------------------------------

F=k*((q

_{1})(q

_{2})/r

^{2})

K = 9E9 Nm

^{2}/c

^{2}

q

_{1}= charge 1 in columbs

q

_{2}= charge 2 in columbs

r = distance between the two charges (in meters)

a

^{2}+ b

^{2}= c

^{2}

sin(xº)=O/H

#

_{1}E#

_{2}= #

_{1}*10^#

_{2}

## The Attempt at a Solution

-----------------------------------------

F

_{1}= 9E9 * ((2.6µC)*( -1.0µC))/.1

^{2}

F

_{1}= -2.34 N

*It's negative because its being pulled down due to attraction

F

_{2}= 9E9 * ((2.6µC)*( 13.0µC))/.224

^{2}

F

_{2}= 5.6 N

**.1m

^{2}+.2m

^{2}=.05 , .05

^{.5}= .224m **

**.1m and hyptonuse angle: sin

^{-1}(.2m/.224) = 63.23º **

** Other angle is 26.76º (90º-63.23º)**

sin(xº)=O/H --> H*sin(xº)=O

5.6 N sin (26.76º) = +2.54 N

+2.54 N - 2.34 N = +.2 N

Is this the correct answer? +.2N

------------------------------------

I tried using 5.6cos(26.56º) = 5.009N (Hcos(xº) = A) for the x forces and it told me I was wrong. I plugged in 5 N & -5 N :|

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