Compare your weight with the attraction (

or repulsion) force of two One Coulomb charges separated one kilometer. Is it larger, smaller or similar?

I already took the subject. It is your turn. We are here to help, not to do all your work for you. A couple of your questions are interesting but most are standard.

I already did them, just looking to check if I did them alright.

Hard for me to tell.

i.imgur(dot)com/WBQnrHY(dot)jpg

Can you tell me if it is alright?
I'm a bit confused about the rise time and the fall time, I don't know if I should consider fall time at all.

I can see it and am looking

I am wondering if the voltage and current should be linear in time upward during the rise and then linear during fall.

In the last one not only are you neutral charge overall but your individual parts tend to be neutral atom by atom. You do not have a concentration of ions in your feet and - ions in your head :)

1 Coulomb charges 1 km apart

E = k Q/r^2
r = 10^3
E = 9*10^9 /10^6 = 9000
EQ = 9000 N

So do u mean my answer in the first question is wrong?

we need to find the total charge passed during the rise and fall

during rise
i = {10^4amps /[10^-5 s]}t
area under that line (the triangle) is in coulombs
Q = .5*10^4*10^-5 = .05 Coulombs

during fall
Q = .5*10^4 *10^-4 = .5 Coulombs

peak power = volts * peak amps
10^7 * 10^4 = 10^11 watts (yikes)

total energy = integral of Vi dt assuming V is constant that is V int i dt
but we know integral i dt = .05 + .5 =
.55
so 10^7 Volts * .55 coulombs total energy

I don't see where the .5^4 comes.

Thanks for the answer but if you could elaborate it would be awesome.

Thnk so much really.

You mean .5 * 10^4 ?

I am saying that current is proportional to time during the rise (and fall)
It goes from 0 to 10^4 in 10^-5 seconds
the area under that straight line is
.5 * 10^4 *10^-5 which is the charge passed in that time
Therefore my answers should be half yours because I do not us the peak but allow for rise time and fall time