Recent content by AdKo

Alright, thanks a lot!

thanks for all your help. I solved by using C=Q/V :smile: that was easy. :biggrin:

I think I totally understand now! Your explanation was great. So would this be correct?? (l_ is the base + height) l_ :1mx1m=1m^2 l_ :100cmx100xcm=10000cm^2

Wait is it like this?: \frac{5cm^2}{1}*\frac{1m}{(100cm)^2} =\frac{5cm^2}{1}*\frac{1m}{10000cm^2} ?? -- I guess that would mean the 5cm^2 already had its square root distibuted? Am I thinking the right way?

By the way, I've always ignored all of the powers that came after the units.. But I guess that's because I never converted them. I really need this explained =/

But isn't cm square also? Or is it like (xm)^2 and you need to distribute so x^2m^2? Cm^2 already has the magnitude distributed right? Is that the reason?

I don't get it. Do you need to count the ^2 too? How do you convert that? I know that 7.6cm=.076m..

My answer is off by 10^-2! Homework Statement An air-filled capacitor consists of two parallel plates, each with an area of 7.60cm^2 and separated by a distance of 1.8mm. If a 20 V potential difference is applied to these plates, calculate the capacitance. Homework Equations...

Homework Statement A parallel-plate capacitor has an area of 5 cm^2, and the plates are separated by 1mm with air between them. The capacitor stores a charge of 400pC. a) What is the potential difference across the plates of the capacitor? b) What is the magnitude of the uniform electric field...

I see, ok thanks for all your help! :smile:

I guess I should make W and F_e positive if I make the acceleration positive? Is that how it works? Wait, how do I know whether the acceleration is upwards or downwards? Haha, I'm getting confused again. =/

I dunno. I have always made forces that pointed downwards negative and the upward forces positive. Which is correct?? Why do you make them positive? I always have trouble deciding this by the way.

Ok, so I found the final speed of the bead without the electrical field by using the equation you gave me: v=\sqrt{2*9.8*5} and the final velocity would be 9.9m/s which is less than the final velocity with the electrical field so that must mean the electrical field points downwards right...

So if there's no drag for in a vacuum, then the only two forces acting in the mass is gravity and the electric force caused by the electric field (F=qE) correct? So how do you even calculate the direction? Are you suppose to use F=ma to find the acceleration? Do you need to use any of the...

Here's my new diagramhttp://img407.imageshack.us/my.php?image=52diagramac0.png Btw, are the forces that act on the bead (when the field is off) its weight and the drag force? How do you calculate its drag force? I'm just confused about that. There would still be a drag force when the electric...