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cause
01-25-2012, 07:14 AM
Hey guys, just working on some Phys 112. I'm definitely rusty at it! My question is this:
http://puu.sh/evZr
and so far I have:
http://puu.sh/evZv
I can't seem to figure out how to solve for the net force...

bluez
04-27-2012, 12:48 PM
this may help.similiar problem= two point charges Q1 and Q2 are held in place 4.50cm apart.

Another point charge -2.50μC of mass 5.5g is initially located 3cm from each of these charges and released from rest.

You observe that the initial acceleration of -2.50μC is 344m/s² upward, parallel to the line connecting the two point charges.

Find Q1 and Q2.

solution=

In order for Q to move upward it must be attracted to q1 and repelled by q2. q1 and q2 must also have the same charge, but opposite signs as just mentioned, so that the perpendicular force will vanish.

Just a reminder, Coulomb's Law, will be needed:
F = kQ1*Q2/r^2
k = 9.0x10^9 Nm^2/C^2

Upward force on Q is (first convert mass from g to kg):
F = ma = (0.0055)(344) = 1.892 N

If Fv is the vertical component of force from q1 then the force from q2 is also Fv. So the total force is 2Fv and it must equal F just calculated.

2Fv = 1.892
Fv = 0.946 N
This is related to the total force between q1 and q by:
cos(A) = Fv/(total force)
A = the angle between an upward vector at Q to the line from Q to q1.
cos(A) = 2.25/3 = 0.75
total force = Fv/cos(A) = 0.946/0.75 = 1.26133 N

Using Coulombs law (stated at the start) and first converting charge from microC to C and distance in cm to m:

1.26133 = (9.0x10^9)[2.5x10^(-6)][q1]/(.03)^2
1.26133 = [(9.0)(2.5)x10^3][q1]/(.03)^2
1.26133 = [(9.0)(2.5)x10^3][q1]/[9x(10^-4)]
1.26133 = [(2.5)x10^7][q1]
q1 = 1.26133 / [(2.5)x10^7]
q1 = 0.5045 x 10^(-7) C

We can the write:
q1 = +0.05045 microC
q2 = -0.05045 microC

Check by seeing if i get the right force between Q and q1:
F = (9.0x10^9)[2.5x10^(-6)][0.05045x10^(-6)]…
F = (9.0x10^9)[2.5x10^(-6)][0.05045x10^(-6)]…
F = [2.5x10][0.05045]
F = 1.26125
And this agrees with the total force calculated above