- #1
I don't understand why ##\Delta V = 2000 V ## in the solutions.
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.
Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.
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.
##q\Delta V = 0.5mv^2 ##
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## ?
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?
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?
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?
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 ##
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?
Oh I though difference meant finial - inital. But I guess initial - finial would also be ok if you took the absolute value.
p.d is a scalar thought, but would a negative sign sort of give it a direction?
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?
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.
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.
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.
That's as magnitude x direction, but as scalar x 1D vector it's ##\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').
