Unit conversion inside a derived gausses law eq

AI Thread Summary
The discussion revolves around converting the unit \(\alpha = \frac{15 \text{ cm} \cdot \text{m}^3}{\mu \text{C}}\) into meters per coulomb. The user struggles with unit conversions, specifically how to handle the centimeter to meter conversion and the microcoulomb to coulomb conversion. They attempt various calculations but are unsure about their results, particularly the expression \(1.5 \times 10^{-7} \text{ m}^4/\text{C}\). The confusion stems from the complexity of the units involved, and the user expresses frustration over the problem's difficulty. Ultimately, the user seeks clarity on the correct conversion method.
bobasp1
Messages
3
Reaction score
0

Homework Statement


Essentially I'm working out a problem and I am given \alpha = \frac{15cm\bullet m^{3}}{\mu C}
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
 
Physics news on Phys.org
15 cm*m^3/micro C* 1Micro C/1X10^6 C *1 m/ 100 cm
 
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
 
and whatsthat m^3 with 15cm
 
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:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top