Point Charges in Square
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
Short
Video
\(\LaTeX\)
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Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Two po charges q_A and q_B are placed at two neighbouring corners of a square. A third po charge q_C is placed at the centre of the square. The values are given in the figure below. center includegraphicswidth.mm#image_path:square# center abcliste abc Calculate the magnitude of the net force acting on the charge q_C. abc Determine the direction of the net force. abcliste
Solution:
abcliste abc The partial forces F_A and F_B on q_C have magnitudes F_A B k_Cfracq_A B q_Cr^ With the distance r given by r frac asqrt fracasqrt we get F_A B k_Cfracq_A B q_Ca^ The two forces are perpicular to each other. We can therefore use the Pythagorean theorem to finde the magnitude of the net force: sscFnet sqrtF_A^+F_B^ FF times kCtimesfracqca^timessqrtqa^+qb^ F approx resultFP- abc The figure below shows how the direction of the net force can be found using the known graphical addition of vectors. The absolute length of the partial force vectors is not relevant. center includegraphicswidth.mm#image_path:squar# center abcliste
Two po charges q_A and q_B are placed at two neighbouring corners of a square. A third po charge q_C is placed at the centre of the square. The values are given in the figure below. center includegraphicswidth.mm#image_path:square# center abcliste abc Calculate the magnitude of the net force acting on the charge q_C. abc Determine the direction of the net force. abcliste
Solution:
abcliste abc The partial forces F_A and F_B on q_C have magnitudes F_A B k_Cfracq_A B q_Cr^ With the distance r given by r frac asqrt fracasqrt we get F_A B k_Cfracq_A B q_Ca^ The two forces are perpicular to each other. We can therefore use the Pythagorean theorem to finde the magnitude of the net force: sscFnet sqrtF_A^+F_B^ FF times kCtimesfracqca^timessqrtqa^+qb^ F approx resultFP- abc The figure below shows how the direction of the net force can be found using the known graphical addition of vectors. The absolute length of the partial force vectors is not relevant. center includegraphicswidth.mm#image_path:squar# center abcliste
Meta Information
Exercise:
Two po charges q_A and q_B are placed at two neighbouring corners of a square. A third po charge q_C is placed at the centre of the square. The values are given in the figure below. center includegraphicswidth.mm#image_path:square# center abcliste abc Calculate the magnitude of the net force acting on the charge q_C. abc Determine the direction of the net force. abcliste
Solution:
abcliste abc The partial forces F_A and F_B on q_C have magnitudes F_A B k_Cfracq_A B q_Cr^ With the distance r given by r frac asqrt fracasqrt we get F_A B k_Cfracq_A B q_Ca^ The two forces are perpicular to each other. We can therefore use the Pythagorean theorem to finde the magnitude of the net force: sscFnet sqrtF_A^+F_B^ FF times kCtimesfracqca^timessqrtqa^+qb^ F approx resultFP- abc The figure below shows how the direction of the net force can be found using the known graphical addition of vectors. The absolute length of the partial force vectors is not relevant. center includegraphicswidth.mm#image_path:squar# center abcliste
Two po charges q_A and q_B are placed at two neighbouring corners of a square. A third po charge q_C is placed at the centre of the square. The values are given in the figure below. center includegraphicswidth.mm#image_path:square# center abcliste abc Calculate the magnitude of the net force acting on the charge q_C. abc Determine the direction of the net force. abcliste
Solution:
abcliste abc The partial forces F_A and F_B on q_C have magnitudes F_A B k_Cfracq_A B q_Cr^ With the distance r given by r frac asqrt fracasqrt we get F_A B k_Cfracq_A B q_Ca^ The two forces are perpicular to each other. We can therefore use the Pythagorean theorem to finde the magnitude of the net force: sscFnet sqrtF_A^+F_B^ FF times kCtimesfracqca^timessqrtqa^+qb^ F approx resultFP- abc The figure below shows how the direction of the net force can be found using the known graphical addition of vectors. The absolute length of the partial force vectors is not relevant. center includegraphicswidth.mm#image_path:squar# center abcliste
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