Unit conversion inside a derived gausses law eq

In summary, the problem involves converting the given value of \alpha = \frac{15cm\bullet m^{3}}{\mu C} into terms of meters/coulombs. The correct solution involves using unit conversions for meters and coulombs, resulting in a value of 1.5e-7 m^4/C. The variable of m^3 with 15cm may be causing confusion, but it is not relevant to the solution.
  • #1
bobasp1
3
0

Homework Statement


Essentially I'm working out a problem and I am given [tex]\alpha = \frac{15cm\bullet m^{3}}{\mu C} [/tex]
and I need to get it in just terms of meters/coulombs

I'm having a real tough time with weird unit conversions like this.

Homework Equations



1m = 100cm 1 micro coulomb = 1 X 10^-6 coulombs

The Attempt at a Solution


1m/100cm * 10^-6 micro c/1c * the above eq. 1.5 x 10^-7 m^4/C pretty sure that isn't correct.Thanks
 
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  • #2
15 cm*m^3/micro C* 1Micro C/1X10^6 C *1 m/ 100 cm
 
  • #3
bobasp1 said:
1m/100cm * 10^-6 micro c/1c * the above eq. 1.5 x 10^-7 m^4/C pretty sure that isn't correct.

Thanks

the 10^-6 of micro coulomb will become 10^6 when it goes up
 
  • #4
and whatsthat m^3 with 15cm
 
  • #5
RTW69 said:
15 cm*m^3/micro C* 1Micro C/1X10^6 C *1 m/ 100 cm
bahh >__< i had it right, I guess my problem isn't correct then Thanks everyone

1.5e-7 m^4/C
cupid.callin said:
and whatsthat m^3 with 15cm

no clue why, its from quest homework service and the problem it self is pretty chill, but that one variable is screwing with me.
 
Last edited:

1. What is unit conversion in the context of Gauss's Law equation?

Unit conversion in the context of Gauss's Law equation refers to the process of converting the units of measurement used in the equation to a more convenient or appropriate unit system. This is often necessary when working with different systems of measurement or when converting between SI units and imperial units.

2. Why is unit conversion important in Gauss's Law equation?

Unit conversion is important in Gauss's Law equation because it allows for consistency and accuracy in calculations. Using the correct units ensures that the equation is being applied correctly and produces meaningful results. It also allows for easier comparison and analysis of data.

3. How do you perform unit conversion in Gauss's Law equation?

To perform unit conversion in Gauss's Law equation, you must first identify the units used in the equation and the desired units. Then, you can use conversion factors or equations to convert from one unit to another. It is important to carefully track the units throughout the calculation to ensure accuracy.

4. Can you give an example of unit conversion in Gauss's Law equation?

Yes, an example of unit conversion in Gauss's Law equation is converting from Coulombs (C) to Statcoulombs (statC) in the equation Q = ε0ΦE. This can be done by using the conversion factor 1 statC = 3.3356 x 10-10 C, resulting in the equation Q = (3.3356 x 10-100ΦE.

5. What are some common units used in Gauss's Law equation?

Some common units used in Gauss's Law equation include meters (m) for distance, Coulombs (C) for charge, Newtons per Coulomb (N/C) for electric field strength, and square meters (m2) for electric flux. These units are typically used in the SI unit system, but other unit systems may also be used depending on the specific application.

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