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.
Isn't Capacity=coulombs/volt ? figure the coulombs of charge.
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
How do you get 28*300 to be 8.0 ?
Generally, we get 7.6 times some power of ten.
oops, 8.6 x some power of ten
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