Calculating Energy Required for Moving Charge: Simple Circuits Problem Solution

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Discussion Overview

The discussion revolves around calculating the energy required to move a charge of 260C through a potential difference of 6V. Participants explore various equations and concepts related to energy, work, and charge in the context of simple circuits.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to calculate energy using the equations for capacitance and energy stored in a capacitor, expressing uncertainty about the correctness of their approach.
  • Another participant suggests using the formula V = W/Q to relate voltage, work, and charge, but notes issues with unit consistency.
  • A later reply clarifies the distinction between work (energy) and power (watts), indicating a misunderstanding in the initial approach.
  • One participant proposes a direct calculation of energy as Energy = Charge * Potential, arriving at a different value of 1560J, and seeks confirmation on this approach.
  • Another participant suggests deriving equations involving voltage, energy, charge, and current, indicating a more complex relationship among these variables.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to the problem, with multiple competing methods and calculations presented. There is no consensus on the final answer or the best method to use.

Contextual Notes

Some participants highlight confusion regarding the use of the symbol W for both work and watts, which may have contributed to misunderstandings in the calculations. The discussion includes various assumptions about the relationships between voltage, charge, and energy without resolving these uncertainties.

Stevecgz
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Homework Statement



Determine the energy required to move 260C through 6V.

Homework Equations



q = Cv
U = (1/2)*C*v^2

The Attempt at a Solution



C = q/v = 260C/6V = 130/3 F
U = (1/2)(130/3 F)(6V)^2 = 780J

I applied the voltage/charge relationship and energy equation for a capacitor because they seemed to be the only equations that made sense with the given values, but I'm not certain this is correct. If someone could let me know if I'm going about this correctly it would be appreciated. Thanks.
 
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I think you need the formula:

V = \frac{W}{Q}

where V = volts, W = work(energy or joules), and Q = coulombs
 
Number2Pencil said:
I think you need the formula:

V = \frac{W}{Q}

where V = volts, W = work(energy or joules), and Q = coulombs

It seems the units don't work out for that formula.

Did you mean V = W/I? If so I think it doesn't help for this problem since I don't know the current and can't find the current from the given values.
 
Ok, I see where I went wrong. I was looking at the variable W as if it had the unit of watts, which in power, not energy. I now see that W is energy is joules, and the formula you wrote does work. Thanks Number2Pencil, I think I can solve the problem now.
 
Stevecgz said:
Ok, I see where I went wrong. I was looking at the variable W as if it had the unit of watts, which in power, not energy. I now see that W is energy is joules, and the formula you wrote does work. Thanks Number2Pencil, I think I can solve the problem now.

A Watt is a Joule/second, if you did the calculation having used Joules in place of Watts, I'd recommend going back and checking your solution over again
 
I think the variable W and W as the symbol for watt caused confusion. So the correct answer would be:

Energy = Charge*Potential = 260C*6V = 1560J

Right?
 
It looks like your going to have to derrive an equation
Try V=E/Q=W/Q... I=Q/t... E=VIt... and Q=Ne

V = voltage in volts E= energy in joules Q= charge W= work in joules I= current in amps N= number of electrons e = electron constant t=time
 
Stevecgz said:
I think the variable W and W as the symbol for watt caused confusion. So the correct answer would be:

Energy = Charge*Potential = 260C*6V = 1560J

Right?

Correct, Steve. Yeah, using W for both work and Watts can cause confusion.
 

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