What is the charge of the 10 billion electrons?

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SUMMARY

The charge of 10 billion electrons is calculated using the formula Q=Ne, where Q is the charge, N is the number of electrons, and e is the elementary charge (1.6E-19 C). Thus, Q = (1E10)(1.6E-19 C) results in a charge of 1.6E-9 C. To determine the energy stored in a 12V battery rated at 35Ah, the power can be calculated using P=IV, leading to energy in Watt-hours, which can then be converted to Joules. The time a battery lasts when supplying a current of 0.5A can be calculated using the relationship 35 A-hr = I*T.

PREREQUISITES
  • Understanding of basic electrical concepts such as charge, current, and voltage.
  • Familiarity with the formula Q=Ne for calculating charge.
  • Knowledge of power calculations using P=IV.
  • Ability to convert between Watt-hours and Joules.
NEXT STEPS
  • Learn about the relationship between charge, current, and time in electrical circuits.
  • Study the concept of energy storage in batteries and how to calculate it.
  • Explore unit conversions between different energy units, such as Watt-hours and Joules.
  • Investigate the implications of battery capacity ratings on circuit performance.
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Students in introductory physics or electrical engineering courses, educators teaching basic electrical concepts, and anyone interested in understanding battery performance and charge calculations.

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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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