# Acceleration of Uranium 238 ion through a potential difference

• Nway
In summary: I'll pass on that. Get on with your homework, is my advice.Are you just thinking that I take absolute value...or did you mean something else?I'll pass on that. Get on with your homework, is my advice.
Nway
Homework Statement
Below.
Relevant Equations
1/2mv^2
I don't understand why the Uranium 238 ions are accelerated

I think ##\Delta V = -2000 V## to accelerate since the ion would be accelerated by more postive charges so ## V_i > V_f ##

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Nway
PeroK said:
I don't understand why ##\Delta V = 2000 V ## in the solutions.

Nway said:
I don't understand why ##\Delta V = 2000 V ## in the solutions.
What do you think it should be?

Nway
PeroK said:
What do you think it should be?
Because ##\Delta V = V_f - V_i ## so if ##\Delta V = 2000 V ## Then ##V_f = 2000V ## which only accelerates if q is negative. But q postive in this question sir.

Nway said:
Because ##\Delta V = V_f - V_i ## so if ##\Delta V = 2000 V ## Then ##V_f = 2000V ## which only accelerates if q is negative.
Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.

Nway and Redbelly98
PeroK said:
Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.
How would this work?

Nway said:
How would this work?
You start the Uranium ion at the appropriate side of the potential difference. Like, you might start a ball at the top of a hill rather than the bottom.

Nway
PeroK said:
You start the Uranium ion at the appropriate side of the potential difference. Like, you might start a ball at the top of a hill rather than the bottom.
##q\Delta V = 0.5mv^2 ##
##v = (\frac {2q\Delta V}{m})^{1/2} ##

However, ##\Delta V = -2000 V## since ##V_i = 2000## (more postive charges so will accelerate the ##U-238## cation) and ##V_f = 0 ##

However since q is positive then this would give a imaginary velocity as there is a negative in numerator. I don't understand why solutions say ##\Delta V = 2000 V##. I think ##\Delta V = -2000 V##

See what I mean?

Nway said:
##q\Delta V = 0.5mv^2 ##
##v = (\frac {2q\Delta V}{m})^{1/2} ##

However, ##\Delta V = -2000 V## since ##V_i = 2000## (more postive charges so will accelerate the ##U-238## cation) and ##V_f = 0 ##

However since q is positive then this would give a imaginary velocity as there is a negative in numerator. I don't understand why solutions say ##\Delta V = 2000 V##. I think ##\Delta V = -2000 V##

See what I mean?
A difference is a magnitude. It doesn't have a sign. The difference between 2 and 3 is the same as the difference between 3 and 2.

Nway
haruspex said:
A difference is a magnitude. It doesn't have a sign. The difference between 2 and 3 is the same as the difference between 3 and 2.
Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?

Nway said:
See what I mean?
If an aparatus has a potential difference of ##2000 \ V##, then positive charges and negative charges may both be accelerated by it (albeit in opposite directions).

Nway
PeroK said:
If an aparatus has a potential difference of ##2000 \ V##, then positive charges and negative charges may both be accelerated by it (albeit in opposite directions).
But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?

Nway said:
But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?
I would say it does and specifying ##-2000 \ V## in the question would be more precise. But, it's clear what the question intends.

Nway
PeroK said:
I would say it does and specifying ##-2000 \ V## in the question would be more precise. But, it's clear what the question intends.
But how can ## - 2000 V## be true since then the velocity will be purely imaginary?

Plugging in numbers:

##v = (\frac {-4000q}{m})^{1/2} ## but q and m is positive so ##v = i(\frac {4000q}{m})^{1/2} ##

PeroK
Nway said:
But how can ## - 2000 V## be true since then the velocity will be purely imaginary?

Plugging in numbers:

##v = (\frac {-4000q}{m})^{1/2} ## but q and m is positive so ##v = i(\frac {4000q}{m})^{1/2} ##
That's completely wrong. If you want to say "the Uranium ion is not accelerated, because the potential difference has the wrong sign", then that is your answer and you can move on to the next question.

Nway
PeroK said:
That's completely wrong. If you want to say "the Uranium ion is not accelerated, because the potential difference has the wrong sign", then that is your answer and you can move on to the next question.
Sorry sir, could you explain a bit more?

Didn't you say that potential difference dose have a sign? And that sign is ##\Delta V = -2000V##?

But @haruspex disagreed.

Nway said:
Sorry sir, could you explain a bit more?

Didn't you say that potential difference dose have a sign? And that sign is ##\Delta V = -2000V##?

But @haruspex disagreed.
I'll pass on that. Get on with your homework, is my advice.

nasu and Nway
PeroK said:
I'll pass on that. Get on with your homework, is my advice.
Are you just thinking that I take absolute value of the p.d? So then ##|\Delta V = - 2000V| = 2000 V ##

However, it seems to me like black magic to take the absolute value to get rid of the - sign and the imaginary velocity.

The question doesn't specify the sign of the potential difference at all. However, if the positive uranium ion started next to the negative electrode, it would stick to it, and not be accelerated, whereas if it started next to the positive electrode, it would be repelled by it and accelerated towards the negative electrode. Since the question does specify that the ion is accelerated, we can conclude that it started next to the positive electrode.
You can also ignore the sign of the potential difference, and compute the amount of kinetic energy it gained from it, wich must be positive, since the ion was accelerated.

Redbelly98, Nway and nasu
Nway said:
Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?
Because that is what "difference" means. It does not specify which way to do the subtraction. What is the difference in height between Everest and Nanga Parbat? Would you answer with a negative value just because I put them in that order?
The question says the ions are accelerated "through a potential difference". It does not say "up through" or "down through", so still does not imply a direction.

PeroK and Nway
haruspex said:
Because that is what "difference" means. It does not specify which way to do the subtraction. What is the difference in height between Everest and Nanga Parbat? Would you answer with a negative value just because I put them in that order?
The question says the ions are accelerated "through a potential difference". It does not say "up through" or "down through", so still does not imply a direction.
Oh I though difference meant finial - inital. But I guess initial - finial would also be ok if you took the absolute value.

haruspex said:
Because that is what "difference" means. It does not specify which way to do the subtraction. What is the difference in height between Everest and Nanga Parbat? Would you answer with a negative value just because I put them in that order?
The question says the ions are accelerated "through a potential difference". It does not say "up through" or "down through", so still does not imply a direction.
p.d is a scalar thought, but would a negative sign sort of give it a direction?

Nway said:
I thought difference meant final - initial.
I would use "change" for that.
Nway said:
p.d is a scalar thought, but would a negative sign sort of give it a direction?
Height is also a scalar, and scalars have direction.

Nway
haruspex said:
I would use "change" for that.

Height is also a scalar, and scalars have direction.
Ain't the difference between scalars and vectors is that scalars have only two possible discrete directions while vectors have COUNTIOUS range of directions?

They normally teach us in school that scalars have magnitude and vectors have magnitude and direction.

Nway said:
Ain't the difference between scalars and vectors is that scalars have only two possible discrete directions while vectors have COUNTIOUS range of directions?
Unless it's a 1D vector space. Some say scalars do form a 1D vector space; that's true except that you have to throw away some structure. E.g. you cannot divide by a vector.
Nway said:
They normally teach us in school that scalars have magnitude and vectors have magnitude and direction.
A common blunder. Scalars, as you note, do have direction, except that, like a 1D vector, it is only a choice of two directions.

Nway
haruspex said:
A common blunder. Scalars, as you note, do have direction, except that, like a 1D vector, it is only a choice of two directions.
I prefer to think that "direction" is a vector attribute only and that scalars have a sign, positive or negative, but not a direction. When I see ##v_x=-3~##m/s, my interpretation is that the x-component of the velocity vector, a scalar, is negative, in whatever way "negative" on the x-axis is defined. In 1D, where there is no need for axes labels, subscript ##x## is commonly dropped but the interpretation remains the same.

I agree about the common blunder. I suspect the confusion arises because of the dropped subscript and the blurring of what is meant by the result. In 1D, the equation ##v=-3~##m/s, can have only one interpretation formally written as ##\mathbf{v}=3~(-\mathbf{\hat x})~##m/s. Thus, ##v=-3~##m/s is meant to be a vector equation but the unit vector is omitted while the negative sign is retained.

If, as you say, scalars have direction, how would you write the scalar in this case to indicate the direction it has? How can you write a scalar equation when your intention is to write a vector equation? If you write ##v=-3~##m/s, then I think that would feed into the confusion. One would interpret ##3~##m/s as the scalar part and the negative sign as the direction part reinforcing the misconception that scalars are always positive. If you write ##v_x=-3~##m/s, then there is no direction involved in ##v_x## because the vector equation is ##\mathbf{v}=v_x~\mathbf{\hat x}## and that's where the direction is.

Nway
kuruman said:
I prefer to think that "direction" is a vector attribute only and that scalars have a sign, positive or negative, but not a direction. When I see ##v_x=-3~##m/s, my interpretation is that the x-component of the velocity vector, a scalar, is negative, in whatever way "negative" on the x-axis is defined. In 1D, where there is no need for axes labels, subscript ##x## is commonly dropped but the interpretation remains the same.
We are so used to thinking of vectors in Cartesian coordinates. But go back to the statement that vectors have "magnitude and direction". That fits more naturally with polar coordinates. In one dimension, the direction consists of the sign.
kuruman said:
I suspect the confusion arises because of the dropped subscript and the blurring of what is meant by the result. In 1D, the equation ##v=-3~##m/s, can have only one interpretation formally written as ##\mathbf{v}=3~(-\mathbf{\hat x})~##m/s.
That's as magnitude x direction, but as scalar x 1D vector it's ##\mathbf{v}=-3~(\mathbf{\hat x})~##m/s.
kuruman said:
One would interpret 3 m/s as the scalar part and the negative sign as the direction part
No, you should interpret it as either magnitude 3, direction -, or as scalar -3 on whatever oriented axis has been specified (like, 'up').

Nway

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