Energy, current and electrical potential of a particle accelerator 🎆

[lpettigrew]

Correct. This is your answer here, everything else that follows is a mess that isn't leading anywhere.
No. As you determined literally one line before, the current is 0.7 A.How can 25 times 10^-9 be smaller than 10^-9?ns/ns wouldn't be per second, so that's again having wrong units. If one bunch come every 25 ns, then how can there be less than one bunch per second?
Ampere per second would be a change of current over time, but the current does not change.

You need to check your answers if they can make sense.

Sorry I think that I had confused myself.

iii. " If one bunch reaches the collision point every 25 ns, what is the average electric current, in Amps, due to these arriving bunches?"

I understand how as you stated previously when you multiply a charge by something per second the product is not a charge but rather a current.
So to answer part iii.
Find how many bunches reach the collision point in one second;
1 second = 1000000000 ns
1000000000ns/25ns= 40000000 = 4.0*10^7 bunches per second
Charge = 1.8423*10^-8 C
Average current = (4.0*10^7 bunches/second)*(1.8423*10^-8 C)
Average current = 0.73692 A

However, does this not neglect the forumla I=Q/t ?
Should current not be 1.8423*10^-8 C/4.0*10^7 bunches/second=4.61*10^-16 A ?

 

[mfb]

t = 25 ns
Or, equivalently:
1/t = 4.0*10⁷ / s

 

[lpettigrew]

t = 25 ns
Or, equivalently:
1/t = 4.0*10⁷ / s

Thank you for your reply. So I=4.0*10^7/ s * 1.8423*10^-8 C = 0.73692 A

However, sorry to ask again but does this not neglect the forumla I=Q/t ?
Should current not be 1.8423*10^-8 C/4.0*10^7 bunches/second=4.61*10^-16 A ?

 

[SammyS]

Thank you for your reply. So I=4.0*10^7/ s * 1.8423*10^-8 C = 0.73692 A

However, sorry to ask again but does this not neglect the forumla I=Q/t ? No.
Should current not be 1.8423*10^-8 C/4.0*10^7 bunches/second=4.61*10^-16 A ? No also.

Each bunch has a charge of 1.8423×10⁻⁸ C and a bunch reaches the collision point in 25ns.
Yes the average, I, is given by I = Q/t . If you want that to be in units of Amps, you need charge to be in Coulombs and time to be in seconds, or some equivalent but strange combination, such as having charge in units of nano-Coulombs and time in nano-seconds. I would convert the 25ns to seconds.
So you could say I = (1.8423×10⁻⁸ C) / (25×10⁻⁹ s) = 0.73692 A .

Alternatively, you can find the total amount of charge reaching the collision point in some other period of time and divide that charge by that time. Essentially, this is what you did - well, what you sort of did - using a time of 1 second. To do this strictly, you have that in 1 second, 4.0×10⁷ bunches reach the collision point. They represent a total charge of 0.73692 C. In this case Q is 0.73692 C and t is 1 s.

By your first calculation quoted in this post, you found the current by multiplying the two following quantities:
"bunches per second" and "Coulombs per bunch". Yes, multiplying these does give the desired current. However, it is not strictly by using I = Q/t .

 

[mfb]

However, sorry to ask again but does this not neglect the forumla I=Q/t ?
Should current not be 1.8423*10^-8 C/4.0*10^7 bunches/second=4.61*10^-16 A ?

You ignored my post (that you quoted) completely.

4.0*10^7 bunches/second is not the time. It's the inverse of the time. 25 ns is the time.