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- Thread starter Nway
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- #2

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I don't understand your question.

- #3

Nway

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I don't understand why ##\Delta V = 2000 V ## in the solutions.I don't understand your question.

- #4

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

- #5

Nway

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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.What do you think it should be?

- #6

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

- #7

Nway

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How would this work?Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.

- #8

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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.How would this work?

- #9

Nway

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##q\Delta V = 0.5mv^2 ##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.

##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?

- #10

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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.##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?

- #11

Nway

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Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?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.

- #12

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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).See what I mean?

- #13

Nway

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But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?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).

- #14

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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 why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?

- #15

Nway

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

Plugging in numbers:

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

- #16

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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.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} ##

- #17

Nway

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Sorry sir, could you explain a bit more?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.

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

But @haruspex disagreed.

- #18

- #19

Nway

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

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

- #20

willem2

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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.

- #21

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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?Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?

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.

- #22

Nway

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Oh I though difference meant finial - inital. But I guess initial - finial would also be ok if you took the absolute value.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.

- #23

Nway

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p.d is a scalar thought, but would a negative sign sort of give it a direction?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.

- #24

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I would use "change" for that.I thought difference meantfinal - initial.

Height is also a scalar, and scalars have direction.p.d is a scalar thought, but would a negative sign sort of give it a direction?

- #25

Nway

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Ain't the difference between scalars and vectors is that scalars have only two possible discrete directions while vectors have COUNTIOUS range of directions?I would use "change" for that.

Height is also a scalar, and scalars have direction.

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

- #26

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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.Ain't the difference between scalars and vectors is that scalars have only two possible discrete directions while vectors have COUNTIOUS range of directions?

A common blunder. Scalars, as you note, do have direction, except that, like a 1D vector, it is only a choice of two directions.They normally teach us in school that scalars have magnitude and vectors have magnitude and direction.

- #27

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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.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 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.

- #28

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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.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.

That's as magnitude x direction, but as scalar x 1D vector it's ##\mathbf{v}=-3~(\mathbf{\hat x})~##m/s.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.

No, you should interpret it as either magnitude 3, direction -, or as scalar -3 on whatever oriented axis has been specified (like, 'up').One would interpret 3 m/s as the scalar part and the negative sign as the direction part

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