Leyden Jar
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:
A simple Leyden jar consists of an acrylic glass tube with length lO and outer radius roO. The inside and outside of the tube are coated with aluminium foils. The Leyden jar has a capacitance of CO. Calculate the thickness of the tube.
Solution:
Using the capacitance of a cylindrical capacitor we have C fracpikappavarepsilon_ Llnfracr_or_i We have to solve this for the inner radius: lnfracr_or_i fracpikappavarepsilon_ LC fracr_or_i expleftfracpikappavarepsilon_ LCright r_i r_oexpleft-fracpikappavarepsilon_ LCright The thickness is therefore given by d r_o-r_i dF rotimesleft-expleft-fracpitimes ktimes ncepstimes lCrightright d approx resultdP- Since the thickness of the tube is quite small compared to the diameter of the tube we can think of the capacitor as a parallel plate capacitor with an area corresponding to the mantle surface of the tube: C kappavarepsilon_fracAd kappavarepsilon_fracpi r_o Ld This can easily be solved for the thickness: d dapprF ktimesncepstimesfracpitimes rotimes lC dappr approx resultdapprP- The result does not perfectly agree with the more complicated calculation for the cylindrical capacitor but it is a good approximation.
A simple Leyden jar consists of an acrylic glass tube with length lO and outer radius roO. The inside and outside of the tube are coated with aluminium foils. The Leyden jar has a capacitance of CO. Calculate the thickness of the tube.
Solution:
Using the capacitance of a cylindrical capacitor we have C fracpikappavarepsilon_ Llnfracr_or_i We have to solve this for the inner radius: lnfracr_or_i fracpikappavarepsilon_ LC fracr_or_i expleftfracpikappavarepsilon_ LCright r_i r_oexpleft-fracpikappavarepsilon_ LCright The thickness is therefore given by d r_o-r_i dF rotimesleft-expleft-fracpitimes ktimes ncepstimes lCrightright d approx resultdP- Since the thickness of the tube is quite small compared to the diameter of the tube we can think of the capacitor as a parallel plate capacitor with an area corresponding to the mantle surface of the tube: C kappavarepsilon_fracAd kappavarepsilon_fracpi r_o Ld This can easily be solved for the thickness: d dapprF ktimesncepstimesfracpitimes rotimes lC dappr approx resultdapprP- The result does not perfectly agree with the more complicated calculation for the cylindrical capacitor but it is a good approximation.
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Exercise:
A simple Leyden jar consists of an acrylic glass tube with length lO and outer radius roO. The inside and outside of the tube are coated with aluminium foils. The Leyden jar has a capacitance of CO. Calculate the thickness of the tube.
Solution:
Using the capacitance of a cylindrical capacitor we have C fracpikappavarepsilon_ Llnfracr_or_i We have to solve this for the inner radius: lnfracr_or_i fracpikappavarepsilon_ LC fracr_or_i expleftfracpikappavarepsilon_ LCright r_i r_oexpleft-fracpikappavarepsilon_ LCright The thickness is therefore given by d r_o-r_i dF rotimesleft-expleft-fracpitimes ktimes ncepstimes lCrightright d approx resultdP- Since the thickness of the tube is quite small compared to the diameter of the tube we can think of the capacitor as a parallel plate capacitor with an area corresponding to the mantle surface of the tube: C kappavarepsilon_fracAd kappavarepsilon_fracpi r_o Ld This can easily be solved for the thickness: d dapprF ktimesncepstimesfracpitimes rotimes lC dappr approx resultdapprP- The result does not perfectly agree with the more complicated calculation for the cylindrical capacitor but it is a good approximation.
A simple Leyden jar consists of an acrylic glass tube with length lO and outer radius roO. The inside and outside of the tube are coated with aluminium foils. The Leyden jar has a capacitance of CO. Calculate the thickness of the tube.
Solution:
Using the capacitance of a cylindrical capacitor we have C fracpikappavarepsilon_ Llnfracr_or_i We have to solve this for the inner radius: lnfracr_or_i fracpikappavarepsilon_ LC fracr_or_i expleftfracpikappavarepsilon_ LCright r_i r_oexpleft-fracpikappavarepsilon_ LCright The thickness is therefore given by d r_o-r_i dF rotimesleft-expleft-fracpitimes ktimes ncepstimes lCrightright d approx resultdP- Since the thickness of the tube is quite small compared to the diameter of the tube we can think of the capacitor as a parallel plate capacitor with an area corresponding to the mantle surface of the tube: C kappavarepsilon_fracAd kappavarepsilon_fracpi r_o Ld This can easily be solved for the thickness: d dapprF ktimesncepstimesfracpitimes rotimes lC dappr approx resultdapprP- The result does not perfectly agree with the more complicated calculation for the cylindrical capacitor but it is a good approximation.
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