Deflection of Charged Pendula
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:
Two balls of mass mO are each susped from a LO long string to form a simple pulum. The two pula are placed such that the centers of the initally neutral balls are at a distance dO. The balls are then charged to QaO and QbO respectively. Determine the horizontal deflection of the two balls. center includegraphicswidth.textwidth#image_path:electric-pula# center H: Derive an for the horizontal displacement. There is no easy way to find a formal solution for this . You can solve it numerically e.g. using a tool like mathematica or graphically by plotting the two sides of the and finding the ersection.
Solution:
includegraphicswidthtextwidth#image_path:doublelectric-pulum-labelled# The vector of the two force vectors vecF_G gravitational force and vecF_C Coulomb force on a ball has to po in the direction of the string. Using similar triangles it follows: fracxy fracF_CF_G fracxsqrtL^-x^ frack_C Q_ Q_m g r^ frack_C Q_ Q_m g d+ x^ This complicated expression for x can be solved numerically e.g. using Mathematica or graphically. For the latter approach you can plot the graphs of the left and right hand side as a function of x in the same diagram using the numerical values for L k_C Q_ Q_ d m and g. The x-value at the ersection po corresponds to the solution. includegraphicswidthtextwidth#image_path:electric-pulum-plot-# The value read from the above diagram is xapprox .cm. A numercical solution using Mathematica leads to a value of x.cmapprox.cm.
Two balls of mass mO are each susped from a LO long string to form a simple pulum. The two pula are placed such that the centers of the initally neutral balls are at a distance dO. The balls are then charged to QaO and QbO respectively. Determine the horizontal deflection of the two balls. center includegraphicswidth.textwidth#image_path:electric-pula# center H: Derive an for the horizontal displacement. There is no easy way to find a formal solution for this . You can solve it numerically e.g. using a tool like mathematica or graphically by plotting the two sides of the and finding the ersection.
Solution:
includegraphicswidthtextwidth#image_path:doublelectric-pulum-labelled# The vector of the two force vectors vecF_G gravitational force and vecF_C Coulomb force on a ball has to po in the direction of the string. Using similar triangles it follows: fracxy fracF_CF_G fracxsqrtL^-x^ frack_C Q_ Q_m g r^ frack_C Q_ Q_m g d+ x^ This complicated expression for x can be solved numerically e.g. using Mathematica or graphically. For the latter approach you can plot the graphs of the left and right hand side as a function of x in the same diagram using the numerical values for L k_C Q_ Q_ d m and g. The x-value at the ersection po corresponds to the solution. includegraphicswidthtextwidth#image_path:electric-pulum-plot-# The value read from the above diagram is xapprox .cm. A numercical solution using Mathematica leads to a value of x.cmapprox.cm.
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Exercise:
Two balls of mass mO are each susped from a LO long string to form a simple pulum. The two pula are placed such that the centers of the initally neutral balls are at a distance dO. The balls are then charged to QaO and QbO respectively. Determine the horizontal deflection of the two balls. center includegraphicswidth.textwidth#image_path:electric-pula# center H: Derive an for the horizontal displacement. There is no easy way to find a formal solution for this . You can solve it numerically e.g. using a tool like mathematica or graphically by plotting the two sides of the and finding the ersection.
Solution:
includegraphicswidthtextwidth#image_path:doublelectric-pulum-labelled# The vector of the two force vectors vecF_G gravitational force and vecF_C Coulomb force on a ball has to po in the direction of the string. Using similar triangles it follows: fracxy fracF_CF_G fracxsqrtL^-x^ frack_C Q_ Q_m g r^ frack_C Q_ Q_m g d+ x^ This complicated expression for x can be solved numerically e.g. using Mathematica or graphically. For the latter approach you can plot the graphs of the left and right hand side as a function of x in the same diagram using the numerical values for L k_C Q_ Q_ d m and g. The x-value at the ersection po corresponds to the solution. includegraphicswidthtextwidth#image_path:electric-pulum-plot-# The value read from the above diagram is xapprox .cm. A numercical solution using Mathematica leads to a value of x.cmapprox.cm.
Two balls of mass mO are each susped from a LO long string to form a simple pulum. The two pula are placed such that the centers of the initally neutral balls are at a distance dO. The balls are then charged to QaO and QbO respectively. Determine the horizontal deflection of the two balls. center includegraphicswidth.textwidth#image_path:electric-pula# center H: Derive an for the horizontal displacement. There is no easy way to find a formal solution for this . You can solve it numerically e.g. using a tool like mathematica or graphically by plotting the two sides of the and finding the ersection.
Solution:
includegraphicswidthtextwidth#image_path:doublelectric-pulum-labelled# The vector of the two force vectors vecF_G gravitational force and vecF_C Coulomb force on a ball has to po in the direction of the string. Using similar triangles it follows: fracxy fracF_CF_G fracxsqrtL^-x^ frack_C Q_ Q_m g r^ frack_C Q_ Q_m g d+ x^ This complicated expression for x can be solved numerically e.g. using Mathematica or graphically. For the latter approach you can plot the graphs of the left and right hand side as a function of x in the same diagram using the numerical values for L k_C Q_ Q_ d m and g. The x-value at the ersection po corresponds to the solution. includegraphicswidthtextwidth#image_path:electric-pulum-plot-# The value read from the above diagram is xapprox .cm. A numercical solution using Mathematica leads to a value of x.cmapprox.cm.
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