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Rotating the Resultant Force on a Charge

  • Thread starter jegues
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  • #1
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Homework Statement



See figure attached for problem statement and figures given

Homework Equations





The Attempt at a Solution



See figure attached for my attempt.

I'm obtaining x^2 = negative in the end so I'm guessing I must have done something wrong, I'm just not quite sure what.

I first found the resultant force on the particle located at the origin.

I let,

[tex]\vec{F_{15}} = z \hat{i}[/tex]

I then looked at resulting x and y components of the resulting force in an attempt to solve for a z which will rotate the resultant force 30 degrees counter clockwise.

After finding z, I set this equal to the "proper" [tex]\vec{F_{15}}[/tex] and solved for the distance, x.

Does anyone see where my mistake is?
 

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Answers and Replies

  • #2
1,097
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Bump, still looking for a little help on this one.

EDIT: Found my mistake, I indicated all the forces in the wrong direction, the negative particles attract the positive one at the origin, not repulse.

Attached my 2nd attempt at the problem, still same dillema.

It makes sense intuitively that we should move charge number one a little closer to charge 5, thus generating the extra force needed to shift the resultant 30 degrees counter clockwise.

I just can't get the math to work out...

EDIT: I think I've worked the source of the problem down to the angle, working things out now...

The problem lies with the angle in my triangle, I am saying it is 60 degrees from the negative x-axis, or equivalently 120 degrees from the positiove x-axis.

tan(120) will produce -root(3) while tan(60) will produce root(3), thus where my missing negative sign was the whole time.

Can't I simply take the angle from the horizontal axis everytime and simply make sign adjustments? How will I know when I need to make those adjustments?
 

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