View attachment 200355
For this question, I got the right answer but I used a different method - I didn't take charge/electron into account.
Q = It = nF
n = (2x10^20)/(6.02x10^23) = 3 x 10^-4
So, I = nF/t ~ (3x10^-4)(10^5)/5 = ~ 6A
Is my method still correct?
For this question, I got the right answer but I used a different method - I didn't take charge/electron into account.
Q = It = nF
n = (2x10^20)/(6.02x10^23) = 3 x 10^-4
So, I = nF/t ~ (3x10^-4)(10^5)/5 = ~ 6A
Is my method still correct?