Solving J.E = W/V with Dimensional Analysis

  • Thread starter AriAstronomer
  • Start date
In summary, Ari is discussing a claim that the current density dotted with the electric field is equal to work per volume. Using dimensional analysis, they keep getting an extra unit of seconds on the left hand side of the denominator. They are trying to figure out what they are doing wrong and ask for help from others. They realize that the claim must be a typo and that it should actually be power per volume.
  • #1
AriAstronomer
48
1
Hey guys,
So I know I'm missing something really stupid, but I can't figure out what it is.
I'm seeing the claim that the current density dotted with the Electric field is equal to Work per volume, or: J.E = W/V. Using dimensional analysis though, I keep getting an extra unit of seconds on the left hand side of the denominator:
(Cm/m^3s)(N/C) = J/(m^3)
Nm/sm^3 = J/m^3
J/sm^3 = J/m^3

What am I doing wrong?
Thanks,
Ari
 
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  • #2
I get the same answer as you using the fact that the electric field has dimensions of voltage / length. Then J*E becomes

[C/(sm2)] * V/m

= (C*J/C)*m-3s-1

= Jm-3s-1

Where exactly are you "seeing this claim?"
 
  • #3
W isn't "work" ,it's work/t(its dimension is J/s),you mistakes at this
 
  • #4
Yeah J dot E is power per unit volume. That's usually the starting point for deriving expressions for the poynting vector and the energy density in the field.
 
  • #5
Ahh Power/Volume. Thanks. I had seen this claim on a flashcard, I'm preparing for GRE's and got a bunch of flash cards sent from a school to help me. Must be a typo on their part then.

Ari
 

FAQ: Solving J.E = W/V with Dimensional Analysis

What is the formula for solving J.E = W/V using dimensional analysis?

The formula for solving J.E = W/V using dimensional analysis is to first determine the units of each variable (J is for joules, E is for energy, W is for work, and V is for volume). Then, use conversion factors to cancel out the units until you are left with the desired unit. For example, if you are solving for energy in joules, you would use the conversion factor 1 J = 1 kg * m^2 / s^2.

Can dimensional analysis be used for any type of equation?

Yes, dimensional analysis can be used for any type of equation as long as it involves converting between units. This method is particularly useful for solving problems in physics and chemistry, but can also be applied to other scientific fields.

What are the benefits of using dimensional analysis to solve equations?

Dimensional analysis helps to ensure that the units in an equation are consistent and accurate. It also allows for a systematic approach to problem solving, making it easier to identify and cancel out units. Additionally, dimensional analysis can help to catch any errors in calculations.

Is it necessary to use dimensional analysis when solving J.E = W/V?

While it is not necessary to use dimensional analysis to solve J.E = W/V, it is a helpful tool for ensuring that the units in the equation are correct and consistent. It can also make the problem-solving process more organized and efficient.

Can dimensional analysis be used to convert between different systems of units?

Yes, dimensional analysis can be used to convert between different systems of units, such as converting from metric units to imperial units. This can be done by using conversion factors between the two systems of units. However, it is important to double check the accuracy of the conversion factors used.

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