Energy, current and electrical potential of a particle accelerator 🎆

[lpettigrew]

Homework Statement

A particle accelerator accelerates bunches of protons, each containing 115 billion particles, to an energy of 6.5 TeV (1 eV is the energy of one proton or electron that is accelerated by a potential of 1 Volt).
i. What is the electrical potential, in Volts, of each particle?
ii. If the proton charge is 1.6 * 10^(-19) C, what is the energy, in Joules, of each bunch?
iii. If one bunch reaches the collision point every 25 ns, what is the average electric current, in Amps, due to these arriving bunches?

Relevant Equations

W=V*Q
I=Q/t

Hello, I have answered the question below but would like some advice on whether I can improve my answer or if anyone is able to check whether I have made any mistakes ?

i. 1 V = 1eV in a 1:1 relationship, therefore;
6.5 TeV = 6.5 TV = 6.5 *10^12V

ii. E=W
W=V * Q
Q=number of particles * charge on each
Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8
W=(6.5 *10^12)*(1.84*10^-8)
W=119,749.5 J

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
The charge on a bunch of protons was earlier shown as 1.8423*10^-8 C.
So Q= ( 4.0 ^ 10^7)*(1.8423*10^-8)=0.73692 C
I=Q/t
I=0.73692/2.5*10^-8 s
I=29476800 A

I think that I have taken a wrong turn on the last question and have confused myself a little confessedly

 

[BvU]

taken a wrong turn

Yeah, you used the 25 ns twice ! Good intuition !

 

[lpettigrew]

Thank you for a reply, how can I correct this? Should it be I=0.73692 A ?

 

[lpettigrew]

Sorry I am confused now

 

[BvU]

I=Q/t
Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8

and t = 25 ns

Of cousre the peak is enormous, but the exercise specifically asks for the average.

 

[lpettigrew]

Thank you for your reply. Oh so,
I=Q/t
I=1.8423*10^-8/2.5*10^-8 s
I=0.73692 A?

Or I=1.8423*10^-8/25

 

[BvU]

If you want Coulomb/second, the choice is rather obvious !

 

[mfb]

I=0.73692 A?

Right.

i. 1 V = 1eV in a 1:1 relationship, therefore;
6.5 TeV = 6.5 TV = 6.5 *10^12V

That's the answer the book expects, but I would like to note that there is no such potential in the LHC. The acceleration is done by alternating fields, not an absurdly large static potential.

Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8
W=(6.5 *10^12)*(1.84*10^-8)

You should work with units consistently. That avoids many mistakes.

 

[lpettigrew]

If you want Coulomb/second, the choice is rather obvious !

Thank you for your reply, so in Amps (coulomb/second)=0.73692 A

 

[lpettigrew]

Right.That's the answer the book expects, but I would like to note that there is no such potential in the LHC. The acceleration is done by alternating fields, not an absurdly large static potential.
You should work with units consistently. That avoids many mistakes.

Thank you for your reply, do you mean
Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8 C
W=(6.5 *10^12)V*(1.84*10^-8)C ?

 

[mfb]

Add an elementary charge in the calculation of Q (currently there is a dimensionless number in the middle) and it looks good.

 

[lpettigrew]

Add an elementary charge in the calculation of Q (currently there is a dimensionless number in the middle) and it looks good.

Sorry, where should I add this (I assume you mean 1.602 x 10^-19)
When calculationing Q, I have already used;
Q=number of particles * charge on each
Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8

 

[mfb]

(115*10^9)*(1.602*10^-19)

That's a multiplication of two real numbers. It doesn't have a unit, so it cannot be a charge.
The elementary charge is 1.602*10^-19 C.

 

[lpettigrew]

That's a multiplication of two real numbers. It doesn't have a unit, so it cannot be a charge.
The elementary charge is 1.602*10^-19 C.

So are you saying to multiply 1.8423*10^-8 by 1.602*10^-19 ? Or to find Q using Q=It?

 

[mfb]

So are you saying to multiply 1.8423*10^-8 by 1.602*10^-19 ?

No, I'm saying you should use units. Multiply 1.8423*10^-8 by the elementary charge, which is 1.602*10^-19 C. The unit matters.

 

[lpettigrew]

So would this be correct ? Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8 C
(The units being coulombs here)

 

[mfb]

That's the same mistake as before. If you write an "=" then both sides should be the same. A number is not a charge.

 

[lpettigrew]

That's the same mistake as before. If you write an "=" then both sides should be the same. A number is not a charge.

I agree, I could just not see a reasonable way to find the charge without a given current. However, could I instead use W=V*Q since work done=energy;
Therefore,
W=6.5 *10^12V * 1.602*10^-19 C
W=0.0000010413 J ~ 1.04 * 10^-6 J (3.s.f)

 

[BvU]

For how many particles is that ?
[edit]never mind - this isn't about the numbers any more

 

[lpettigrew]

I assume that what I had written earlier was wrong, since this was only considering the charge on a single proton, not the 115 billion in the bunch?

So, should I find the charge on the bunch of protons by muliplying the charge on a single prootn (1.602*10^-19) by the number of protons in a bunch (115 billion)
(115*10^9)*(1.602*10^-19)=1.8423*10^-8 (would the units of this be coulombs or could they not be since 115 billion is just a meausre of particles and does not have a unit?)

Then since E=V*Q
E=(6.5 *10^12)*(1.84*10^-8)
E=119,749.5 J

I know that I keep reiterating the same mistakes but I do not know how to correct my answer?

 

[BvU]

Hi,

We hammer on these units because they are important: checking the dimension helps find mistakes and wrong formulas.

To the extreme: numbers indeed do not have a dimemsion, but for scale factors you can simply add them (silently):
Energy (per bunch) = Energy (per proton) * (protons /bunch) =
$\qquad\qquad$ Charge (per proton) * Potential * (protons /bunch)

So

So, should I find the charge on the bunch of protons by muliplying the charge on a single prootn (1.602*10^-19 C ) by the number of protons in a bunch (115 billion protons/bunch )
(115*10^9 protons/bunch)*(1.602*10^-19 C/proton )=1.8423*10^-8 C/bunch )

(would the units of this be coulombs or could they not be since 115 billion is just a meausre of particles and does not have a unit?)

Then since E=V*Q
E=(6.5 *10^12 V)*(1.84*10^-8 C/bunch ) = 119749.5 J/bunch = 120 kJ/bunch

and the answer ends up as e.g. :

The energy per bunch is 120 kJ​

After a bit of practice (how useful, these exercises !) this all comes naturally .

Side note: dealing with energy in eV or in J is just a matter of a dimensionless conversion factor

1.602*10^-19 Coulomb/e x 1 J/(Coulomb Volt)​

 

[lpettigrew]

Hi,

We hammer on these units because they are important: checking the dimension helps find mistakes and wrong formulas.

To the extreme: numbers indeed do not have a dimemsion, but for scale factors you can simply add them (silently):
Energy (per bunch) = Energy (per proton) * (protons /bunch) =
$\qquad\qquad$ Charge (per proton) * Potential * (protons /bunch)

SoAfter a bit of practice (how useful, these exercises !) this all comes naturally .

Side note: dealing with energy in eV or in J is just a matter of a dimensionless conversion factor

1.602*10^-19 Coulomb/e x 1 J/(Coulomb Volt)​

Oh, I see what you mean now! Yes, I understand and will make use of these units. Can I ask, are my other answers to parts 1 and 3 correct (or do I have underlying mistkaes also?) Thank you for your help

 

[SammyS]

Oh, I see what you mean now! Yes, I understand and will make use of these units. Can I ask, are my other answers to parts 1 and 3 correct (or do I have underlying mistakes also?) Thank you for your help

At this point in the thread, I can't tell, at present, what your answers are to parts 1 and 3. Since the issues raised above by @mfb and @BvU emphasize using units consistently and correctly, if you repeat the answers below, please be complete.

 

[lpettigrew]

At this point in the thread, I can't tell, at present, what your answers are to parts 1 and 3. Since the issues raised above by @mfb and @BvU emphasize using units consistently and correctly, if you repeat the answers below, please be complete.

Oh of course sorry, I have attached them below:

1 V = 1eV in a 1:1 relationship, therefore;
6.5 TeV = 6.5 TV = 6.5 *10^12V

ii. E=W
W=V * Q
Energy (per bunch) = Energy (per proton) * (protons /bunch)
Energy (per bunch) = Charge (per proton) * Potential * (protons /bunch)
E=(115*10^9 protons/bunch)*(1.602*10^-19 C/proton )*(6.5 *10^12 V)
W=119,749.5 J ~ 120MJ

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
The charge on a bunch of protons was earlier shown as 1.8423*10^-8 C.
So Q= ( 4.0 ^ 10^7)*(1.8423*10^-8)=0.73692 C
I=Q/t
I=0.73692/2.5*10^-8 s
I=29476800 A

 

[SammyS]

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
The charge on a bunch of protons was earlier shown as 1.8423*10^-8 C.
So Q= ( 4.0 ^ 10^7)*(1.8423*10^-8)=0.73692 C
I=Q/t
I=0.73692/2.5*10^-8 s
I=29476800 A

Parts i and ii look satisfactory.

Part iii has a problem.

Either, consider the amount of charge reaching the collision 'point' in 25 nanoseconds , (1 bunch worth)

OR

consider the amount of charge reaching the collision 'point' in 1 second.

Added in Edit:
And for the line in your post which I bolded :

You should have included the units with the number, 1.8423×10^{− 8}
Q= ( 4.0 ^ 10^7)*(1.8423*10^-8 C)=0.73692 C

 

[lpettigrew]

Parts i and ii look satisfactory.

Part iii has a problem.

Either, consider the amount of charge reaching the collision 'point' in 25 nanoseconds , (1 bunch worth)

OR

consider the amount of charge reaching the collision 'point' in 1 second.

Added in Edit:
And for the line in your post which I bolded :

You should have included the units with the number, 1.8423×10^{− 8}
Q= ( 4.0 ^ 10^7)*(1.8423*10^-8 C)=0.73692 C

Sorry I must have forgotten to include the units again. Would;
Q= ( 4.0 ^ 10^7)*(1.8423*10^-8C)=0.73692 C
I=Q/t
I=0.73692C/2.5*10^-8 s
I=29476800 A

not be the amount of charge reaching the collision point every second?

And would the amount of charge reaching the collision 'point' in 25 nanoseconds being one bunch be;
I=Q/t
I=(1.8423*10^-8C)/25ns
I=7.3692*10^-10 A

(but I do not think that my second suggestion is anywhere near correct, at least not to arrive at the units of amps?)

 

[mfb]

Q= ( 4.0 ^ 10^7)*(1.8423*10^-8C)=0.73692 C

You are repeating an earlier mistake that would be easy to spot if you would use units.
4*10⁷ bunches per second. If you multiply a charge by something per second the product is not a charge any more. It's a current. And if you do that correctly you get the correct answer, as e.g. I confirmed in post 8.

I=29476800 A
I=7.3692*10^-10 A

These two are not correct. The second one is missing that 1 ns is 10^-9 s.

 

[lpettigrew]

You are repeating an earlier mistake that would be easy to spot if you would use units.
4*10⁷ bunches per second. If you multiply a charge by something per second the product is not a charge any more. It's a current. And if you do that correctly you get the correct answer, as e.g. I confirmed in post 8.
These two are not correct. The second one is missing that 1 ns is 10^-9 s.

Thank you for your reply.

iii. Oh so would;
I= ( 4.0 ^ 10^7 bunches/second)*(1.8423*10^-8 coulombs)=0.73692 A
So the current passed per seocnd is 7.3692*10^-10 A?

Or to find the charge passed per bunch would I use; Q=energy/voltage
Q=120,000 J /6.5 *10^12 V
Q=1.846154×10^−8 C

Then I=Q/t
If 1 ns = 1*10^-9 s
Then 25 ns = 2.5*10^-10 seconds
I=1.846154×10^−8 C/2.5*10^-10 s
I=73.84616 A per bunch ~ 74 A per bunch

There are 1*10^-9 ns/25ns= = 4.0*10^-11 bunches per second
To find the current per second 74 * 4.0*10^-11 bunches per second = 2.96×10^-9 A per second

Would this be right because the question asks
"If one bunch reaches the collision point every 25 ns, what is the average electric current, in Amps, due to these arriving bunches?"

One ampere is defined as a flow of charge at the rate of 1 coulomb per second, so should the caluclated answer be the bunches per second, not per ns? That is why I multiplied the current per bunch by the number of bunches per second.

 

[lpettigrew]

Sorry have I completely muddled my answer up? I think I have confused myself now

 

[mfb]

I= ( 4.0 ^ 10^7 bunches/second)*(1.8423*10^-8 coulombs)=0.73692 A

Correct. This is your answer here, everything else that follows is a mess that isn't leading anywhere.

So the current passed per seocnd is 7.3692*10^-10 A?

No. As you determined literally one line before, the current is 0.7 A.

Then 25 ns = 2.5*10^-10 seconds

How can 25 times 10^-9 be smaller than 10^-9?

There are 1*10^-9 ns/25ns= = 4.0*10^-11 bunches per second

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?

To find the current per second 74 * 4.0*10^-11 bunches per second = 2.96×10^-9 A 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.

 

[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.