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Right.
Oh oh Gracy, you really have a knack for threads that stretch seemingly forever
Take this from #17:
I seem to sense that you have less inhibition with the alternative form: If the force is ##q\vec E##, the work needed to move a charge over a differential ## d\vec r## is ##dW = - q\vec E \cdot d\vec r## . In terms of potential ##dV = - \vec E \cdot d\vec r##
Then:
And:
Gradually the notion "there is a sentence" in your problem statement evolves into something like "Is the following statement correct:
Finally:
Your post #28 shows you have indeed developed mastery of the subject at hand. #27 had some (small) room for improvement: ##
W/q = \Delta V = -\vec E\cdot\Delta \vec r ## , with ##
\Delta \vec r = \vec b-\vec a## would have avoided your confusion. There is only a slight visual difference between a \cdot that stands for the inner product of two vectors and a period (which is a bit lower on the line). (*)
But:
Since your occupation still states "pedantic student", I am glad I still have room to offer for improvement in post #28: ##
dV = \int_A^B \vec E {\bf \cdot } d\vec r\ ## should be ##\Delta V = \int_A^B dV = \int_A^B -\vec E {\bf \cdot } d\vec r\ ##
And if you really want it perfect you also replace ##\vec E. \vec b - \vec a\ ## by ##\vec E\cdot \left (\vec b - \vec a\ \right ) ## (so: \cdot and brackets) !
As we know: nobody's perfect...
(*) some pedantic but well meant advice:
Oh oh Gracy, you really have a knack for threads that stretch seemingly forever
Take this from #17:
I think it is clear to you by now that the derivative wrt a vector has components. Because from a point ##\vec r## you can go in different directions ##d\vec r##. Dividing by a vector (and ##d\vec r## is a vector) isn't defined. I know you dislike calculus, but we really can't do without. So browse (or better: study carefully) here and here . And when doing an exercise, make a full stop at a points where you can't -- at least in principle -- distinguish what a derivative or an integral in the expression entails. I grant you it's sometimes difficult to imagine (I, for one, still have that with curl, in spite of all the explanatory examples).gracy said:
I seem to sense that you have less inhibition with the alternative form: If the force is ##q\vec E##, the work needed to move a charge over a differential ## d\vec r## is ##dW = - q\vec E \cdot d\vec r## . In terms of potential ##dV = - \vec E \cdot d\vec r##
This is in fact still a differential form (i first called this the integral form, but that is after you add the ##\int## left and right).
Then:
If the sentence says "...VA is more than VB ... " then that really means you are supposed to say something about the potential difference.
And:
Gradually the notion "there is a sentence" in your problem statement evolves into something like "Is the following statement correct:
In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m) "
So why didn't you render the problem statement a bit more faithfully when starting the thread ?Finally:
Your post #28 shows you have indeed developed mastery of the subject at hand. #27 had some (small) room for improvement: ##
W/q = \Delta V = -\vec E\cdot\Delta \vec r ## , with ##
\Delta \vec r = \vec b-\vec a## would have avoided your confusion. There is only a slight visual difference between a \cdot that stands for the inner product of two vectors and a period (which is a bit lower on the line). (*)
But:
Since your occupation still states "pedantic student", I am glad I still have room to offer for improvement in post #28: ##
dV = \int_A^B \vec E {\bf \cdot } d\vec r\ ## should be ##\Delta V = \int_A^B dV = \int_A^B -\vec E {\bf \cdot } d\vec r\ ##
And if you really want it perfect you also replace ##\vec E. \vec b - \vec a\ ## by ##\vec E\cdot \left (\vec b - \vec a\ \right ) ## (so: \cdot and brackets) !
As we know: nobody's perfect...
(*) some pedantic but well meant advice:
- use right mouse button | show Math as ... | TeX commands a bit more to evolve TeX skills
- keep math equations left and right sides together inside one single ##\#\# ## ... ## \#\# ## block: the = should be a TeX equals sign ##=## , not a text equals sign.. Same for plus (+ vs ##+##) and minus (- vs ##-##).