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I don't understand why ##\Delta V = 2000 V ## in the solutions.PeroK said:I don't understand your question.
What do you think it should be?Nway said:I don't understand why ##\Delta V = 2000 V ## in the solutions.
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.PeroK said:What do you think it should be?
Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.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.
How would this work?PeroK said:Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.
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 said:How would this work?
##q\Delta V = 0.5mv^2 ##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.
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 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?
Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?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.
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 said:See what I mean?
But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?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).
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 said:But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?
But how can ## - 2000 V## be true since then the velocity will be purely imaginary?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.
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 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} ##
Sorry sir, could you explain a bit more?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.
Are you just thinking that I take absolute value of the p.d? So then ##|\Delta V = - 2000V| = 2000 V ##PeroK said:I'll pass on that. Get on with your homework, is my advice.
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?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## ?
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?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.
I would use "change" for that.Nway said:I thought difference meant final - initial.
Height is also a scalar, and scalars have direction.Nway said:p.d is a scalar thought, but would a negative sign sort of give it a 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?haruspex said:I would use "change" for that.
Height is also a scalar, and scalars have direction.
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: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.Nway said:They normally teach us in school that scalars have magnitude and vectors have magnitude and direction.
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.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.
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 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.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.
No, you should interpret it as either magnitude 3, direction -, or as scalar -3 on whatever oriented axis has been specified (like, 'up').kuruman said:One would interpret 3 m/s as the scalar part and the negative sign as the direction part