Charged ring and point charge
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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\(\LaTeX\)
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Exercise:
Derive a formal expression for the force between a uniformly charged ring and a po charge on the ring's central axis. center includegraphicswidth.mm#image_path:charged-ring-# center
Solution:
We divide the ring o a large number N segments of equal length with charge dQsscQring/N. The force dF of one such segment on the po charge q is shown in the figure. center includegraphicswidth.mm#image_path:charged-ring-solution-# center The same amount of charge dQ'dQ on the opposite side of the ring exerts a force dF' on the po charge. The combined force sscdFnet is given by fracsscdFnet/dF fracdr Longrightarrow sscdFnet dF fracdr k_C fracdQ qr^ fracdr k_C fracsscQring q dN r^ k_C fracsscQring q dN leftR^ + d^right^/ There are N/ such pairs of opposite segments i.e. the total force on q is sscFnet fracN sscdFnet fracN k_C fracsscQring q dN leftR^ + d^right^/ k_C fracsscQring q dleftR^ + d^right^/ Remark: Coulomb's force law only yields the correct force between two po charges i.e. in the limit N to infty. Because of the symmetry of the charge distribution on the ring the correct result can be derived using segments of finite length.
Derive a formal expression for the force between a uniformly charged ring and a po charge on the ring's central axis. center includegraphicswidth.mm#image_path:charged-ring-# center
Solution:
We divide the ring o a large number N segments of equal length with charge dQsscQring/N. The force dF of one such segment on the po charge q is shown in the figure. center includegraphicswidth.mm#image_path:charged-ring-solution-# center The same amount of charge dQ'dQ on the opposite side of the ring exerts a force dF' on the po charge. The combined force sscdFnet is given by fracsscdFnet/dF fracdr Longrightarrow sscdFnet dF fracdr k_C fracdQ qr^ fracdr k_C fracsscQring q dN r^ k_C fracsscQring q dN leftR^ + d^right^/ There are N/ such pairs of opposite segments i.e. the total force on q is sscFnet fracN sscdFnet fracN k_C fracsscQring q dN leftR^ + d^right^/ k_C fracsscQring q dleftR^ + d^right^/ Remark: Coulomb's force law only yields the correct force between two po charges i.e. in the limit N to infty. Because of the symmetry of the charge distribution on the ring the correct result can be derived using segments of finite length.
Meta Information
Exercise:
Derive a formal expression for the force between a uniformly charged ring and a po charge on the ring's central axis. center includegraphicswidth.mm#image_path:charged-ring-# center
Solution:
We divide the ring o a large number N segments of equal length with charge dQsscQring/N. The force dF of one such segment on the po charge q is shown in the figure. center includegraphicswidth.mm#image_path:charged-ring-solution-# center The same amount of charge dQ'dQ on the opposite side of the ring exerts a force dF' on the po charge. The combined force sscdFnet is given by fracsscdFnet/dF fracdr Longrightarrow sscdFnet dF fracdr k_C fracdQ qr^ fracdr k_C fracsscQring q dN r^ k_C fracsscQring q dN leftR^ + d^right^/ There are N/ such pairs of opposite segments i.e. the total force on q is sscFnet fracN sscdFnet fracN k_C fracsscQring q dN leftR^ + d^right^/ k_C fracsscQring q dleftR^ + d^right^/ Remark: Coulomb's force law only yields the correct force between two po charges i.e. in the limit N to infty. Because of the symmetry of the charge distribution on the ring the correct result can be derived using segments of finite length.
Derive a formal expression for the force between a uniformly charged ring and a po charge on the ring's central axis. center includegraphicswidth.mm#image_path:charged-ring-# center
Solution:
We divide the ring o a large number N segments of equal length with charge dQsscQring/N. The force dF of one such segment on the po charge q is shown in the figure. center includegraphicswidth.mm#image_path:charged-ring-solution-# center The same amount of charge dQ'dQ on the opposite side of the ring exerts a force dF' on the po charge. The combined force sscdFnet is given by fracsscdFnet/dF fracdr Longrightarrow sscdFnet dF fracdr k_C fracdQ qr^ fracdr k_C fracsscQring q dN r^ k_C fracsscQring q dN leftR^ + d^right^/ There are N/ such pairs of opposite segments i.e. the total force on q is sscFnet fracN sscdFnet fracN k_C fracsscQring q dN leftR^ + d^right^/ k_C fracsscQring q dleftR^ + d^right^/ Remark: Coulomb's force law only yields the correct force between two po charges i.e. in the limit N to infty. Because of the symmetry of the charge distribution on the ring the correct result can be derived using segments of finite length.
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