What is the charge of the 10 billion electrons?

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Homework Help Overview

The discussion revolves around a battery and lightbulb circuit, specifically focusing on calculating the charge of ten billion electrons passing through a wire. The context includes concepts from introductory physics, such as charge, current, and energy storage in batteries.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the equation Q=Ne to find the charge of electrons and question whether this approach is sufficient. There are discussions about the implications of using Amp-hours for energy calculations and how to relate power to energy.

Discussion Status

Several participants have offered guidance on the calculations, with some suggesting that the original poster's approach is valid while others emphasize the importance of unit conversions. The conversation has shifted towards energy calculations and the relationship between current and battery capacity.

Contextual Notes

Participants note the importance of stating units in calculations, and there is a focus on understanding how to apply Amp-hours in the context of battery life and energy storage. Some participants express uncertainty about the methods used and seek clarification on the relationships between current, power, and energy.

Marcin H
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Homework Statement

:[/B]
I have a battery and lightbulb circuit (in series). 12V battery rated at 35Ah.

Ten billion electrons pass a planar cross section of the wire. What is the charge of the 10 billion electrons?

Homework Equations


Q=Ne
I=dq/dt
1A=1C/s

The Attempt at a Solution


Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.

Q = Ne
Q=(#electrons)(electron charge)
Q= (1E10)(1.6E-19 C)
Q = 1.6 E -9 C
 
Last edited:
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Not an advanced physics problem, so moved to Intro Physics
 
Marcin H said:
Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.
Assuming you have the question right, that's all there is to it. (Get the sign right.)
 
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Doc Al said:
Assuming you have the question right, that's all there is to it. (Get the sign right.)
Ok. Follow up question. How many Joules of energy does the battery I mention store? I would use .5CV^2 or .5QV, but this is what I learned in E&M physics, so I think there is another way. This is EGR 110 intro to electrical and computer engineering, so there are no pre-recs. Is there a way to do this using unit conversions?
 
Marcin H said:

Homework Statement

:[/B]
I have a battery and lightbulb circuit (in series). 12V battery rated at 35Ah.

Ten billion electrons pass a planar cross section of the wire. What is the charge of the 10 billion electrons?

Homework Equations


Q=Ne
I=dq/dt
1A=1C/s

The Attempt at a Solution


Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.

Q = Ne
Q=(#electrons)(electron charge)
Q= (1E10)(1.6E-19)
Q = 1.6 E -9

Always state your units; chances are the question would be marked wrong without them. Otherwise, it looks OK.
 
Marcin H said:
How many Joules of energy does the battery I mention store?
Hint: If the battery delivered a current I, what power would it be delivering?
 
Doc Al said:
Hint: If the battery delivered a current I, what power would it be delivering?
Yeah I was thinking about using P=IV, but I'm not sure we can. Can we still use that if we are given current in Amp hours?
12V battery rated at 35Ah.
 
Marcin H said:
Yeah I was thinking about using P=IV, but I'm not sure we can. Can we still use that if we are given current in Amp hours?
12V battery rated at 35Ah.
Using P = VI is the first step. Now how do you go from power to energy? (Don't plug in numbers yet, just play with the equations symbolically.)
 
Doc Al said:
Using P = VI is the first step. Now how do you go from power to energy? (Don't plug in numbers yet, just play with the equations symbolically.)
Oh I see. P=IV will give you units of WattHours and then you just do unit conversions and get it to Joules. Got it! Thanks!
 
  • #10
Marcin H said:
P=IV will give you units of WattHours
Power will be in units of Watts; if you multiply by the time, then you'll have energy.
 
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  • #11
Doc Al said:
Power will be in units of Watts; if you multiply by the time, then you'll have energy.
Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)
 
  • #12
Marcin H said:
Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)
Good.
 
  • #13
Doc Al said:
Good.
Ran into another problem. Assume the battery is supplying .5A of current to a light bulb. How long will the battery last? (Battery is 12V rated at 35Ah) I am supposed to use the 35Ah to find the time or do I have to somehow use the .5A. I don' t think there is a way to find how long it will last with just Amps. You need Amp hours to do this right?
 
  • #14
Marcin H said:
I am supposed to use the 35Ah to find the time or do I have to somehow use the .5A.
You need to use both Amp-Hours and Amps to solve for the time.
 
  • #15
Doc Al said:
You need to use both Amp-Hours and Amps to solve for the time.
Ok so, I used P=IV to find that the battery supplies 6W. I can do the same thing to find that it provides 420Wh. So can I just use unit conversions to find the hours? 6W/420Wh. Watts cancel and you are left with 1/70h. So 70 hours? Is this correct?
 
  • #16
Marcin H said:
Ok so, I used P=IV to find that the battery supplies 6W. I can do the same thing to find that it provides 420Wh. So can I just use unit conversions to find the hours? 6W/420Wh. Watts cancel and you are left with 1/70h. So 70 hours? Is this correct?
Sure.

All you have to do is realize that the ampere-hours are given so: 35 A-hr = I*T = .5A*T. Solve for T and get 70 hours.
 
  • #17
Doc Al said:
Sure.

All you have to do is realize that the ampere-hours are given so: 35 A-hr = I*T = .5A*T. Solve for T and get 70 hours.
Can you explain that a bit more? Where did you get the "35 A-hr = I*T = .5A*T." part from
 
  • #18
The battery capacity, expressed in terms of Amp-Hours, equals Current*Time. Draw a lower current and the battery can run for a longer time.
 
  • #19
35 A-hr = (.5 A)*(70 hr) = (1 A)*(35 hr) = ... etc.
 
  • #20
Ok, that makes sense. More sense than what I did. Is the way I did it a bad way to go about this? I don't know if I would have noticed to use that equation BC=AT on a test or quiz. I was just looking at units and trying to get hours through unit conversion.
 
  • #21
Marcin H said:
Is the way I did it a bad way to go about this?
No, just a bit more work than needed. Don't think of it as "unit conversion" though.
 

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