Finding the charge of a drum, i got the steps i think

  • Thread starter Thread starter mr_coffee
  • Start date Start date
  • Tags Tags
    Charge Drum
Click For Summary

Homework Help Overview

The discussion revolves around calculating the total charge on the drum of a photocopying machine, given its dimensions and the electric field just above its surface. The problem involves understanding charge density and its relation to electric fields, specifically in the context of a charged cylinder.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between electric field and charge density, questioning the interpretation of charge density as area versus linear charge density. There are attempts to recalculate using corrected dimensions and formulas.

Discussion Status

Some participants have provided guidance on recalculating values and interpreting charge density correctly. There is an ongoing exploration of different interpretations and calculations, with some participants expressing confusion about the concepts involved.

Contextual Notes

Participants note discrepancies in the dimensions used for calculations, highlighting the importance of accurate unit conversion. The discussion also references a follow-up problem involving changes to the drum's dimensions while maintaining the electric field.

mr_coffee
Messages
1,613
Reaction score
1
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^-10 Which is wrong. The book got .32x10^-6C. Any idea where i messsed up at?
 
Physics news on Phys.org
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^-10 Which is wrong. The book got .32x10^-6C. Any idea where i messsed up at?
for a charged cylinder, Gauss's Law yields:

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

where δ is the charge per unit length. Recalculate your results using the proper δ interpretation. also, the drum length is 0.42 m, and the radius R is 0.06 m. your values for these dimensions are off by a factor of 10!
 
So i can't use [tex]E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R}[/tex] to find the linear charge?
 
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^-10 Which is wrong. The book got .32x10^-6C. Any idea where i messsed up at?
mr_coffee said:
So i can't use
[tex]E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R}[/tex]
to find the linear charge?
for a (long) charged cylinder, you CAN and SHOULD use
[tex]E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R}[/tex]
to determine charge per unit LENGTH (δ) when given values for the other variables...
the problem is that you interpreted δ to be an area charge density in your original solution (see RED terms in your original solution above).

recalculate your results with the proper "linear charge density" interpretation of δ, and also correct your unit errors (see Msg #2), and you'll get the textbook answer.
 
Last edited:
I'm confused on what else I could do...
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. :confused:
 
mr_coffee said:
I'm confused on what else I could do...
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. :confused:
(charge per unit LENGTH of cylinder) = δ = (charge on cylinder)/(LENGTH of cylinder)
 
Thank you~ I finally got part A. PArt b says, The manufacturer wishes to produce a desktop version of the machine. This requires reducing the drunm length to 28cm and the diameter to 8.0cm. The e field at the drum surface must not change. What must be the charge on this new drum? So i used the same formula:

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

Similar threads

How do I solve a surface charge problem for a cylindrical drum?
  • · Replies 2 ·
Replies
2
Views
2K