Asked by Anonymous

Two 4.0cm × 4.0cm square aluminum electrodes, spaced 0.50mm apart are connected to a 300V battery.

What is the charge on the positive electrode?

I already found the capacitance which turned out to be 28 pF.
Answered by bobpursley
Isn't Capacity=coulombs/volt ? figure the coulombs of charge.
Answered by Anonymous
yes! that's exactly what I thought. So i ended up figuring that out and got 2.8*10^-9 coulombs but somehow it's still wrong
Answered by bobpursley
How can 300*28Pico be 2.8 times some power? It has to be 7.6 as the numerals.

Answered by Anonymous
This is my full work...

First I had to find the capacitance so,

C=[(8.85*10^-12)(.04)^2]/.0005
=2.832*10^-11 F
=28.32 pF

Now I have to find the charge on the positive electrode, so i tried

Q=CV
=(28*10^-12)(300V)
=8.0*10^-9

but it's wrong
Answered by bobpursley
How do you get 28*300 to be 8.0 ?

Generally, we get 7.6 times some power of ten.
Answered by bobpursley
oops, 8.6 x some power of ten
Answered by Anonymous
I have no idea, that's what my calculator is coming up as when i type in
(28*10^-12)(300)

and since my homework is asking for 2 sig figs, i put 8.0
Answered by Anonymous
are you sure 8.6?? i'm getting 8.4
Answered by Anonymous
nvm bob, i figured it out and got it right on my homework thanks for your help!
Answered by Anonymous
add a pi
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