|Feb6-06, 07:37 PM||#1|
charged mass on a string
A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet of infinite extent. The acceleration of gravity is 9.8 m/s^2 and the permittivity of free space is 8.85 x 10^-12 C^2/Nm^2. Find the angle [tex] \theta [/tex] the thread makes with the vertically charge sheet. Answer in units of degrees.
mass of ball= 1 g
Areal charge density of the sheet= 0.23 [tex] \mu C/m^2[/tex]
length of the string = 78.9 cm
Then force of charge= qE= q[tex] \sigma [/tex] / 2E_0
We did some of this problem in class and went through the long process of drawing a free body diagram and summing up the components, we found that it was easier to use the pythagorean theorem to solve for T.
I found that T= [tex] \sqrt (mg)^2 + (qE)^2 [/tex]
So T= [tex] \sqrt 96.04 + 1.32 x 10^-5 [/tex]
So T= 9.8.
Then I plugged it into what we got for the forces in the y-direction, which was [tex] \theta= cos^-1 (-mg/T) [/tex]
So theta= cos ^-1 (-9.8/9.8)
= 180 degrees which is wrong... can someone help me please?
|Feb6-06, 07:52 PM||#2|
Check your calculation of [itex]q E[/itex] and [itex]mg[/itex] (note that m = 0.001 kg). What's [itex]q[/itex]?
|Feb6-06, 10:23 PM||#3|
Well I stupidly forgot to change to kg, but I'm still getting the wrong answer.
[tex] T= \sqrt (mg)^2 + (qE)^2 [/tex]
mg= .001 * 9.8 = .0098
qE= [tex] q \sigma/2 E_o [/tex]
qE= 2.8 x 10^-7 * 2.3 x 10^-7 / 2 * 8.85 x 10^-12
T= [tex] \sqrt (.0098)^2 + (.00364)^2 [/tex]
[tex] \theta= cos^-1 (-mg/T) [/tex]
[tex] \theta = cos^-1 (-.0098/.0104) [/tex]
[tex] \theta= 159 degrees [/tex]
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