Fill Level
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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Exercise:
A cylindrical tank consists of conducting plates at the top and bottom and instulating side walls. It is partially filled with a liquid with dielectric constant kappa. The fill level can be determined by measuring the capacitance between the top and bottom plate. abcliste abc Derive a formal expression for the fill level as a fraction of the height of the tank as a function of the measured capacitance C and the capacitance of the empty tank C_. abc The empty tank has a capacitance CeO and is hO high. It is filled with benzene. The measured capacitance is CmO. Calculate the fill level. abcliste
Solution:
abcliste abc The partially filled tank can be erpreted as two capacitors in series. Their capacitances are C_ kappavarepsilon_fracAx h C_ varepsilon_fracA-x h where x is the fraction of the tank height filled with the liquid. It follows for the series capacitance sscCs leftfracC_+fracC_right^- leftfracx hkappavarepsilon_ A+frac-x hvarepsilon_ Aright^- C_leftfracxkappa+-xright^- C_leftfracx+kappa-xkapparight^- C_leftfracx-kappa+kappakapparight^- C_frackappax-kappa+kappa We have to solve this expression for x: sscCsleftx-kappa+kapparight C_kappa C_s x-kappa C_-C_skappa x resultfracsscCs-C_sscCsfrackappakappa- We can verify the solution for the extreme cases sscCsC_ empty tank and sscCskappa C_ full tank: sscxmin fracC_-C_C_frackappakappa- sscxmax frackappa C_-C_kappa C_frackappakappa- frackappa-kappafrackappakappa- Both results are as expected. abc The dielectric constant of benzene is kO. Using the solution from a we can express the fill level h_f as follows: h_f x h hfF fracCm-CeCmtimesfrackk-times h hf approx resulthfS abcliste
A cylindrical tank consists of conducting plates at the top and bottom and instulating side walls. It is partially filled with a liquid with dielectric constant kappa. The fill level can be determined by measuring the capacitance between the top and bottom plate. abcliste abc Derive a formal expression for the fill level as a fraction of the height of the tank as a function of the measured capacitance C and the capacitance of the empty tank C_. abc The empty tank has a capacitance CeO and is hO high. It is filled with benzene. The measured capacitance is CmO. Calculate the fill level. abcliste
Solution:
abcliste abc The partially filled tank can be erpreted as two capacitors in series. Their capacitances are C_ kappavarepsilon_fracAx h C_ varepsilon_fracA-x h where x is the fraction of the tank height filled with the liquid. It follows for the series capacitance sscCs leftfracC_+fracC_right^- leftfracx hkappavarepsilon_ A+frac-x hvarepsilon_ Aright^- C_leftfracxkappa+-xright^- C_leftfracx+kappa-xkapparight^- C_leftfracx-kappa+kappakapparight^- C_frackappax-kappa+kappa We have to solve this expression for x: sscCsleftx-kappa+kapparight C_kappa C_s x-kappa C_-C_skappa x resultfracsscCs-C_sscCsfrackappakappa- We can verify the solution for the extreme cases sscCsC_ empty tank and sscCskappa C_ full tank: sscxmin fracC_-C_C_frackappakappa- sscxmax frackappa C_-C_kappa C_frackappakappa- frackappa-kappafrackappakappa- Both results are as expected. abc The dielectric constant of benzene is kO. Using the solution from a we can express the fill level h_f as follows: h_f x h hfF fracCm-CeCmtimesfrackk-times h hf approx resulthfS abcliste
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Exercise:
A cylindrical tank consists of conducting plates at the top and bottom and instulating side walls. It is partially filled with a liquid with dielectric constant kappa. The fill level can be determined by measuring the capacitance between the top and bottom plate. abcliste abc Derive a formal expression for the fill level as a fraction of the height of the tank as a function of the measured capacitance C and the capacitance of the empty tank C_. abc The empty tank has a capacitance CeO and is hO high. It is filled with benzene. The measured capacitance is CmO. Calculate the fill level. abcliste
Solution:
abcliste abc The partially filled tank can be erpreted as two capacitors in series. Their capacitances are C_ kappavarepsilon_fracAx h C_ varepsilon_fracA-x h where x is the fraction of the tank height filled with the liquid. It follows for the series capacitance sscCs leftfracC_+fracC_right^- leftfracx hkappavarepsilon_ A+frac-x hvarepsilon_ Aright^- C_leftfracxkappa+-xright^- C_leftfracx+kappa-xkapparight^- C_leftfracx-kappa+kappakapparight^- C_frackappax-kappa+kappa We have to solve this expression for x: sscCsleftx-kappa+kapparight C_kappa C_s x-kappa C_-C_skappa x resultfracsscCs-C_sscCsfrackappakappa- We can verify the solution for the extreme cases sscCsC_ empty tank and sscCskappa C_ full tank: sscxmin fracC_-C_C_frackappakappa- sscxmax frackappa C_-C_kappa C_frackappakappa- frackappa-kappafrackappakappa- Both results are as expected. abc The dielectric constant of benzene is kO. Using the solution from a we can express the fill level h_f as follows: h_f x h hfF fracCm-CeCmtimesfrackk-times h hf approx resulthfS abcliste
A cylindrical tank consists of conducting plates at the top and bottom and instulating side walls. It is partially filled with a liquid with dielectric constant kappa. The fill level can be determined by measuring the capacitance between the top and bottom plate. abcliste abc Derive a formal expression for the fill level as a fraction of the height of the tank as a function of the measured capacitance C and the capacitance of the empty tank C_. abc The empty tank has a capacitance CeO and is hO high. It is filled with benzene. The measured capacitance is CmO. Calculate the fill level. abcliste
Solution:
abcliste abc The partially filled tank can be erpreted as two capacitors in series. Their capacitances are C_ kappavarepsilon_fracAx h C_ varepsilon_fracA-x h where x is the fraction of the tank height filled with the liquid. It follows for the series capacitance sscCs leftfracC_+fracC_right^- leftfracx hkappavarepsilon_ A+frac-x hvarepsilon_ Aright^- C_leftfracxkappa+-xright^- C_leftfracx+kappa-xkapparight^- C_leftfracx-kappa+kappakapparight^- C_frackappax-kappa+kappa We have to solve this expression for x: sscCsleftx-kappa+kapparight C_kappa C_s x-kappa C_-C_skappa x resultfracsscCs-C_sscCsfrackappakappa- We can verify the solution for the extreme cases sscCsC_ empty tank and sscCskappa C_ full tank: sscxmin fracC_-C_C_frackappakappa- sscxmax frackappa C_-C_kappa C_frackappakappa- frackappa-kappafrackappakappa- Both results are as expected. abc The dielectric constant of benzene is kO. Using the solution from a we can express the fill level h_f as follows: h_f x h hfF fracCm-CeCmtimesfrackk-times h hf approx resulthfS abcliste
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