A small nonconducting ball of mass m = 6.4 g and charge q =8.00×10 -8 C (distrib

Question

A small nonconducting ball of mass m = 6.4g and charge q =8.00×10-8C(distributed uniformly throughits volume) hangs from an insulating thread that makes an angle = 20° with a vertical, uniformly charged nonconductingsheet (shown in cross section).
Considering the weight of the ball and assuming that the sheetextends far vertically and into and out of the page, calculate thesurface charge density of the sheet.
I tried working this one out, and here is the way I didit: F = m*sin(20) = 2.19 N
F = EQ E = F/Q = 27375000 N/C
For a single nonconducting sheet E = /(2e0) where ischarge density and permittivity e0 = 8.8542E-12 F/m.
Thus, = 2e0*E =2*8.8542E-12*27375000 = 0.0004845375 C/m^2
That SEEMS right to me, but it's not. I think I'm missing aconversion somewhere! :(
Oh, and if this is figured out, this can be used for Chapter23, Page 41 in the labeled book since there is no reference on itas of now! :)
I tried working this one out, and here is the way I didit: F = m*sin(20) = 2.19 N
F = EQ E = F/Q = 27375000 N/C
For a single nonconducting sheet E = /(2e0) where ischarge density and permittivity e0 = 8.8542E-12 F/m.
Thus, = 2e0*E =2*8.8542E-12*27375000 = 0.0004845375 C/m^2
That SEEMS right to me, but it's not. I think I'm missing aconversion somewhere! :(
Oh, and if this is figured out, this can be used for Chapter23, Page 41 in the labeled book since there is no reference on itas of now! :)

Explanation / Answer

Hey, I must be in your class. I still have the 3 true falses - Ihate those. Anyways, you did everything right, except for finding F. You used F=mg*sin(20), you should have used F=mg*tan(20). Alsoremember that your mass needs to be converted to kg. So your forceequation should be: F=.0064*9.81*tan(20)=.0229 N Do everything else the same way you did, and that should be all.Good luck!

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