Magnitude and Direction of a charge interaction

themonk
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Homework Statement


26.P39.jpg

What is the direction of the force [tex]\vec{F}[/tex] on the -10 nC charge in the figure? Give your answer as an angle measured cw from the +x-axis. Problem 26.39 in Physics for Scientists and Engineers Second Edition by Knight
Magnitude of the force is 4.3E-3 N

Homework Equations



As far as I know, this is the only relevant equations:

[tex]\vec{F}[/tex]=[tex]k*q_{1}*q_{2}/d^{2}[/tex]

The Attempt at a Solution


I already found the magnitude as listed above. But I need to find [tex]\theta[/tex] in the clockwise direction.

The force of A (being the -5 nC charge) I figured was direction -4.3E-4 [tex]\hat{j}[/tex] and -1.28E-3 [tex]\hat{i}[/tex] and the other charge had a force of -4.5E-3 I double checked with an answer from someone else (MasteringPhysics also said it was correct). I figured I would use a trigonometric function for the final part, ie the angle measured in the clockwise direction. In lab we used tangent, but was confused as to why (tangent is opposite over adjacent):

tan([tex]\theta[/tex])=4.07E-3/1.28E-3 ==> [tex]\theta[/tex]=tan[tex]^{-1}[/tex](3.179)

which is 72.54 degrees. I added 180 to it to get 252.54, but apparently that is not correct (I don't completely understand the cw and ccw part).

Is there any line of thought that I should proceed with?

(sorry about my English)
 
Last edited:
on Phys.org
Show the coordinate axes and the forces in your figure, please.

ehild
 
I can't edit the problem, but the x-axis ([tex]\hat{i}[/tex]) is left-right and the y-axis ([tex]\hat{j}[/tex]) would be up-down, if I am not mistaken. The origin would then be at the -10nC.
 
In this case the force on the -10 nC charge from the -5 nC one has only y component, and that form the 15 nC charge has both x and y components. Check your calculation and text.

ehild
 

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