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

In summary, the drum of the photocpying machine in problem 16 has a length of 42cm and a diameter of 12cm. The charge on the drum can be calculated using the formula \delta = q/Surface Area, where δ is the charge density, q is the charge, and the surface area is given by 2PIrh. However, in the original solution, the charge density was interpreted as an area charge density instead of a linear charge density. This led to incorrect results, which can be corrected by using the proper interpretation of δ and correcting unit errors. In part B, the manufacturer needs to reduce the drum length to 28cm and diameter to 8.0cm while maintaining the same E field at
  • #1
mr_coffee
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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?
 
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  • #2
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!
 
  • #3
So i can't use [tex] E \ = \ \frac{\delta}{2 \pi \epsilon_{0} R} [/tex] to find the linear charge?
 
  • #4
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.
 
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  • #5
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:
 
  • #6
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)
 
  • #7
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:?
 
  • #8
mr_coffee said:
Solved for [tex] \delta [/tex] = 8.142x10^-8.
Redo that calculation.
 
  • #9
Thanks! worked now!
 

1. How do I find the charge of a drum?

To find the charge of a drum, you will need to follow a few steps. First, determine the type of drum you have and locate any markings or labels that may indicate the charge. If there are no markings, you can use a multimeter to measure the voltage of the drum. Finally, use the formula Q=CV to calculate the charge, where Q is the charge, C is the capacitance of the drum, and V is the voltage measured by the multimeter.

2. What is the purpose of finding the charge of a drum?

Knowing the charge of a drum is important for understanding its capacity and potential energy. This information can also be useful for properly grounding a drum and avoiding any electric shocks or accidents.

3. Can I find the charge of a drum without a multimeter?

While using a multimeter is the most accurate way to find the charge of a drum, it is possible to estimate the charge by comparing the drum to a known capacitor with a known charge. However, this method may not be as precise.

4. What are the units for charge and voltage when finding the charge of a drum?

The units for charge are typically measured in coulombs (C) and the units for voltage are measured in volts (V).

5. Is it dangerous to find the charge of a drum?

Finding the charge of a drum can be dangerous if proper precautions are not taken. It is important to use a multimeter correctly and to avoid touching any exposed wires or terminals. If you are unsure or uncomfortable with finding the charge of a drum, it is best to seek assistance from a trained professional.

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