- #1

mr_coffee

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**1.21x10^-10**Which is wrong. The book got .32x10^-6C. Any idea where i messsed up at?

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- Thread starter mr_coffee
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- #1

mr_coffee

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- #2

geosonel

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for a chargedmr_coffee said:1.21x10^-10Which is wrong. The book got .32x10^-6C. Any idea where i messsed up at?

[tex] E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} [/tex]

where δ is the

- #3

mr_coffee

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So i can't use [tex] E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} [/tex] to find the linear charge?

- #4

geosonel

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mr_coffee said:The drum of the photocpying machine in problem 16 has a length of 42cm and a diameter of 12cm. What is the total charge on the drum? In problem 16 it says, The e filed just above the surface of the charged drum of a photocopying machine has a magnitude E of 2.3x10^5 N/C. So I first found the charge density [itex]\delta[/itex]. E = [itex]\delta[/itex]/2PIEor; [itex]\delta[/itex] = E(2PI)(8.85x10^-12)(.006m) = 7.67x10^-8; Then i used [itex]\delta[/itex] = q/Surface Area. q = (2PIrh)[itex]\delta[/itex]; q = 2PI(.006m)(.042m)*7.67x10^-8 =1.21x10^-10Which is wrong. The book got .32x10^-6C. Any idea where i messsed up at?

for a (long) charged cylinder, you CAN and SHOULD usemr_coffee said:So i can't use

[tex] E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} [/tex]

to find the linear charge?

[tex] E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} [/tex]

to determine

the problem is that you interpreted δ to be an area charge density in your original solution (see

recalculate your results with the proper "linear charge density" interpretation of

Last edited:

- #5

mr_coffee

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you said, "the problem is that you interpreted δ to be an area charge density in your original solution" How else can u interpret it? I only know [itex] \delta [/itex] = q/Surface Area. I don't see how else u can find the charge without that formula and solve for q.

- #6

geosonel

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(charge per unitmr_coffee said:

you said, "the problem is that you interpreted δ to be an area charge density in your original solution" How else can u interpret it? I only know [itex] \delta [/itex] = q/Surface Area. I don't see how else u can find the charge without that formula and solve for q.

- #7

mr_coffee

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[tex] E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} [/tex]

Solved for [tex] \delta [/tex] = 8.142x10^-8. I used r = .04m, I kept E the same. Then i used [tex]\delta[/tex] = q/length; q = (8.142x10^-8)(.28m); q = 2.3x10^-8, it should be .14x10^-6C. Any idea where i mseed up:?

- #8

Doc Al

Mentor

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Redo that calculation.mr_coffee said:Solved for [tex] \delta [/tex] = 8.142x10^-8.

- #9

mr_coffee

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Thanks! worked now!

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