Charges/Coulomb's Law/Electric Fields

calcuseless

1. Homework Statement
The picture shows a small 2g mass suspended from a massless thread. When a horizontal electric field of 2000 N/C pointing left to right is applied, the mass moves to the position shown and then comes to a stop. Calculate the magnitude of the charge on the mass.

http://imgur.com/3uSDc [Broken]

2. Homework Equations
Coulomb's Law: F= (k*q1*q2)/r2
F=qE

3. The Attempt at a Solution
I know the charge is negative by the picture, but I can't use these equations without more information. Am I missing an equation I should know or just a concept that lets me ignore something?

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Delphi51

Homework Helper
If there were no gravity, the charged mass would be pushed straight out. It hangs at an angle dependent of the strength of the electric force AND the gravitational force on the mass. Those forces are are in perpendicular directions, so you have a 2D problem to deal with. Begin with a sketch of all the forces and write that the sum of the forces is zero in the horizontal direction, and also in the vertical direction. Usually in these 2D problems you have to work with the two equations together as a system to find the answer.

collinsmark

Homework Helper
Gold Member
Hello calcuseless,
I know the charge is negative by the picture, but I can't use these equations without more information. Am I missing an equation I should know or just a concept that lets me ignore something?
Yes, also consider the force of gravity, and the tension on the string.

Draw a free body diagram. You'll have 3 forces acting on the mass. The electric force, gravity and the tension. You can break up the tension into its components and solve the the answer.

[Edit Delphi51 beat me to the post. Serves me right for getting a cup of coffee in the middle of posting. ]

calcuseless

Ty - Fg = 0
Ty = m * g
sin(40°) = m * g
sin(40°) = .002kg * 9.8N

Tx - FE = 0
Tx = q * E
cos(40°) = q * E
cos(40°) = q * 2000N/C
q = cos(40°) / 2000N/C
q = 3.83 * 10-4 Coulombs

Is this correct?

Should Tx = m * g * cos(40°) instead?

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Delphi51

Homework Helper
sin(40°) = m * g
should be T*sin(40°) = m * g
Same for the horizontal.

Since T appears in both equations, you'll have to solve one of them for T and then substitute the expression into the other equation to eliminate T. I have a much smaller answer for q.

collinsmark

Homework Helper
Gold Member
Should Tx = m * g * cos(40°) instead?
No, not quite. You need to break the tension up, not the force of gravity!
should be T*sin(40°) = m * g
That's not quite right either. Correct, you need to put the T in there multiplied by the trig function, but care needs to be taken about using the correct trig function.

Take a look at where the angle is (this time it is not with respect to the x-axis, but rather with respect to the y-axis). Remember that

sinθ = opposite/hypotenuse

(And eventually, [you'll need this later] tanθ = sinθ/cosθ = opposite/adjacent)

collinsmark

Homework Helper
Gold Member
And something that might be worth double checking, in addition to the above post(s):

In the OP and in the figure, m = 2g. If that is what the problem statement is, than that is what it is. But it just strikes me as a little weird. It would make more common sense to me if the mass was 2/g. As in m = (2 [N])/g. Before giving your final answer, it might be worth double checking the original problem statement and verifying for sure that it states "m = 2g."

Delphi51

Homework Helper
Thank you very much for catching the sin/cos error, collinsmark!
I read the mass "2g" to be 2 grams . . .

collinsmark

Homework Helper
Gold Member
Thank you very much for catching the sin/cos error, collinsmark!
I read the mass "2g" to be 2 grams . . .
Grams! That's it. Never-mind my previous post. Grams.

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