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Doorframe problem

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Abelard
#1
Jan9-11, 01:58 PM
P: 33
A doorframe is twice as tall as it is wide. There is a positive charge on the top left corner and an equal but negative charge in the top right corner. What is the direction of the electric force due to these charges on a negatively charged dust mite in the bottom left corner of the doorframe?

So I know the how to divide into components and analyze in the x an y directions to figure out the net force in x an y direction. But the problem is the values I plugged in were not correct because it was wrong. my answer was 32 or so degree.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Abelard
#2
Jan9-11, 04:05 PM
P: 33
Any brave souls?
gneill
#3
Jan9-11, 04:45 PM
Mentor
P: 11,614
Maybe show a little more of your attempt?

Abelard
#4
Jan9-11, 09:35 PM
P: 33
Doorframe problem

So I assumed that the positive charge at the top left corner is 1.602e-19C, an elementary charge. Likewise, the negative charge at the top right corner is assumed to be -1.602e-19C. The dust mite, although unspecified of its value, is negatively charged, so I assumed its charge to be -1.602e-19C. So the attractive electrostatic force between the mite and the positively charged particle is (0, 5.768e-29)=(Fx,Fy) assuming that the length along the side is 2m and the width is 1m. For the repulsive force between the mite and the negative charge, the angle made by the diagonal line, connecting the mite and the charge, and the length of the doorframe is arcsin(1/sqrt(5)) or 26.565degrees. The net electrostatic force between the negatively charged particle and the mite is 5.768e-29 N. So, using trigonometry, I get (Fx,Fy)=(-2.5795e-29,-5.159e-29). Then summing the forces, (Fx,Fy)=(-2.5795e-29, 6.0895e-30). Theta is then, atan(6.0895e-30/-2.5795e-29)+360= 346.717degrees. Any flaw in this?
gneill
#5
Jan9-11, 10:07 PM
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P: 11,614
Quote Quote by Abelard View Post
The net electrostatic force between the negatively charged particle and the mite is 5.768e-29 N. So, using trigonometry, I get (Fx,Fy)=(-2.5795e-29,-5.159e-29). Then summing the forces, (Fx,Fy)=(-2.5795e-29, 6.0895e-30). Theta is then, atan(6.0895e-30/-2.5795e-29)+360= 346.717degrees. Any flaw in this?
I think you should recheck the force due to the negative charge at the top right corner. You may have used the length of the door frame instead of the diagonal. Also, estimate the quadrant that the resultant should fall in by looking at your diagram; make sure that your calculated angle for the resultant agrees.
Abelard
#6
Jan10-11, 04:29 PM
P: 33
I really don't have an intuitive sense as to where the resultant electrostatic force will point to. So all I can rely on is calculation. So the right angle is 320.3868? The angle is actually atan(1/sqrt(5)). So using the angle, 24.095degrees, the ordered set of x and y component of electrostatic force vector is (-1.88e-29,-4.212e-29). So adding up all the x and y forces, the resultant force will have (-1.88e-29,1.556e-29). So taking arctangent;atan(1.556e-29/-1.88e-29)+360= 320.38degrees since it's in the fourth quadrant.
gneill
#7
Jan10-11, 04:54 PM
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P: 11,614
Quote Quote by Abelard View Post
I really don't have an intuitive sense as to where the resultant electrostatic force will point to. So all I can rely on is calculation. So the right angle is 320.3868? The angle is actually atan(1/sqrt(5)). So using the angle, 24.095degrees, the ordered set of x and y component of electrostatic force vector is (-1.88e-29,-4.212e-29). So adding up all the x and y forces, the resultant force will have (-1.88e-29,1.556e-29). So taking arctangent;atan(1.556e-29/-1.88e-29)+360= 320.38degrees since it's in the fourth quadrant.
The distance from the top right corner to the bottom left is greater than the distance from the top left corner to the bottom left (the diagonal is longer than the side). Since all charges have the same magnitude you expect the magnitude of the force with the shorter separation to be larger. Or, if you prefer, the one with the larger distance between them will be smaller.

The force between charges always lies along the line joining the charges.

So, if you were to sketch the forces on the charge at the lower left you'd have one pointing along the direction of the diagonal to the lower left (repulsive), and one, slightly larger, pointing straight up towards the upper left corner. If you "add" these vectors graphically on your sketch, where does the resultant end up?


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