Point Dipole
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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Exercise:
A po dipole is an idealisation where we ase that the two charges are arbitrarily close together i.e. that the distance d between them is much smaller than the distance r to a po in the field. In this case the electric field vector is given by vecEvec r frack_Cr^ left leftvecp hatrright hatr - vecpright where vecpq vecd is the dipole moment vecr the displacement vector from the centre of the dipole to the po in the field and hatr a unit vector in the direction of vecr i.e. hatrvecr/r. abcliste abc Derive the expressions for the electric field at a po on the line defined by the direction of the dipole i.e. vecr parallel vecp and for the field at a po perpicular to the dipole i.e. vecr perp vecp. abc Calculate the components of the electric field produced by a water molecule dipole moment ppWO for a distance rO and an angle thO between the direction of the dipole moment and the displacement vector. abcliste
Solution:
abcliste abc For a po on the line defined by the direction of the dipole we know that vecp hatr p and p hatr vecp It follows for the electric field vector vecE frack_Cr^ left vecp-vecp right EparF For a po at a right angle to the dipole the dot product is zero. Therefore the electric field vector is vecE EperpF abc We ase that the dipole moment is oriented in the positive x direction: vecp pmatrix p pmatrix The unit vector hatr can then be written as hatr fracvecrhatr fracr pmatrix cosvartheta sinvartheta pmatrixr pmatrix cosvartheta sinvartheta pmatrix The dot product is then just vecp hatr p_x hatr_x + p_y hatr_y + p_z hatr_z p cosvartheta It follows for the components of the field vector E_x frack_Cr^ left p cosvartheta hatr_x - p_x right frack_Cr^ left p cosvartheta hatr_x - p right ExF frack times pWleftrright^ times left times cos^th - right Ex approx resultExP E_y frack_Cr^ left p cosvartheta hatr_y - p_y right frack_Cr^ left p cosvartheta hatr_y - right EyF frack times pWleftrright^ times left times costh times sinth right Ey approx resultEyP E_z frack_Cr^ left p cosvartheta hatr_z - p_z right frack_Cr^ left p cosvartheta - right abcliste
A po dipole is an idealisation where we ase that the two charges are arbitrarily close together i.e. that the distance d between them is much smaller than the distance r to a po in the field. In this case the electric field vector is given by vecEvec r frack_Cr^ left leftvecp hatrright hatr - vecpright where vecpq vecd is the dipole moment vecr the displacement vector from the centre of the dipole to the po in the field and hatr a unit vector in the direction of vecr i.e. hatrvecr/r. abcliste abc Derive the expressions for the electric field at a po on the line defined by the direction of the dipole i.e. vecr parallel vecp and for the field at a po perpicular to the dipole i.e. vecr perp vecp. abc Calculate the components of the electric field produced by a water molecule dipole moment ppWO for a distance rO and an angle thO between the direction of the dipole moment and the displacement vector. abcliste
Solution:
abcliste abc For a po on the line defined by the direction of the dipole we know that vecp hatr p and p hatr vecp It follows for the electric field vector vecE frack_Cr^ left vecp-vecp right EparF For a po at a right angle to the dipole the dot product is zero. Therefore the electric field vector is vecE EperpF abc We ase that the dipole moment is oriented in the positive x direction: vecp pmatrix p pmatrix The unit vector hatr can then be written as hatr fracvecrhatr fracr pmatrix cosvartheta sinvartheta pmatrixr pmatrix cosvartheta sinvartheta pmatrix The dot product is then just vecp hatr p_x hatr_x + p_y hatr_y + p_z hatr_z p cosvartheta It follows for the components of the field vector E_x frack_Cr^ left p cosvartheta hatr_x - p_x right frack_Cr^ left p cosvartheta hatr_x - p right ExF frack times pWleftrright^ times left times cos^th - right Ex approx resultExP E_y frack_Cr^ left p cosvartheta hatr_y - p_y right frack_Cr^ left p cosvartheta hatr_y - right EyF frack times pWleftrright^ times left times costh times sinth right Ey approx resultEyP E_z frack_Cr^ left p cosvartheta hatr_z - p_z right frack_Cr^ left p cosvartheta - right abcliste
Meta Information
Exercise:
A po dipole is an idealisation where we ase that the two charges are arbitrarily close together i.e. that the distance d between them is much smaller than the distance r to a po in the field. In this case the electric field vector is given by vecEvec r frack_Cr^ left leftvecp hatrright hatr - vecpright where vecpq vecd is the dipole moment vecr the displacement vector from the centre of the dipole to the po in the field and hatr a unit vector in the direction of vecr i.e. hatrvecr/r. abcliste abc Derive the expressions for the electric field at a po on the line defined by the direction of the dipole i.e. vecr parallel vecp and for the field at a po perpicular to the dipole i.e. vecr perp vecp. abc Calculate the components of the electric field produced by a water molecule dipole moment ppWO for a distance rO and an angle thO between the direction of the dipole moment and the displacement vector. abcliste
Solution:
abcliste abc For a po on the line defined by the direction of the dipole we know that vecp hatr p and p hatr vecp It follows for the electric field vector vecE frack_Cr^ left vecp-vecp right EparF For a po at a right angle to the dipole the dot product is zero. Therefore the electric field vector is vecE EperpF abc We ase that the dipole moment is oriented in the positive x direction: vecp pmatrix p pmatrix The unit vector hatr can then be written as hatr fracvecrhatr fracr pmatrix cosvartheta sinvartheta pmatrixr pmatrix cosvartheta sinvartheta pmatrix The dot product is then just vecp hatr p_x hatr_x + p_y hatr_y + p_z hatr_z p cosvartheta It follows for the components of the field vector E_x frack_Cr^ left p cosvartheta hatr_x - p_x right frack_Cr^ left p cosvartheta hatr_x - p right ExF frack times pWleftrright^ times left times cos^th - right Ex approx resultExP E_y frack_Cr^ left p cosvartheta hatr_y - p_y right frack_Cr^ left p cosvartheta hatr_y - right EyF frack times pWleftrright^ times left times costh times sinth right Ey approx resultEyP E_z frack_Cr^ left p cosvartheta hatr_z - p_z right frack_Cr^ left p cosvartheta - right abcliste
A po dipole is an idealisation where we ase that the two charges are arbitrarily close together i.e. that the distance d between them is much smaller than the distance r to a po in the field. In this case the electric field vector is given by vecEvec r frack_Cr^ left leftvecp hatrright hatr - vecpright where vecpq vecd is the dipole moment vecr the displacement vector from the centre of the dipole to the po in the field and hatr a unit vector in the direction of vecr i.e. hatrvecr/r. abcliste abc Derive the expressions for the electric field at a po on the line defined by the direction of the dipole i.e. vecr parallel vecp and for the field at a po perpicular to the dipole i.e. vecr perp vecp. abc Calculate the components of the electric field produced by a water molecule dipole moment ppWO for a distance rO and an angle thO between the direction of the dipole moment and the displacement vector. abcliste
Solution:
abcliste abc For a po on the line defined by the direction of the dipole we know that vecp hatr p and p hatr vecp It follows for the electric field vector vecE frack_Cr^ left vecp-vecp right EparF For a po at a right angle to the dipole the dot product is zero. Therefore the electric field vector is vecE EperpF abc We ase that the dipole moment is oriented in the positive x direction: vecp pmatrix p pmatrix The unit vector hatr can then be written as hatr fracvecrhatr fracr pmatrix cosvartheta sinvartheta pmatrixr pmatrix cosvartheta sinvartheta pmatrix The dot product is then just vecp hatr p_x hatr_x + p_y hatr_y + p_z hatr_z p cosvartheta It follows for the components of the field vector E_x frack_Cr^ left p cosvartheta hatr_x - p_x right frack_Cr^ left p cosvartheta hatr_x - p right ExF frack times pWleftrright^ times left times cos^th - right Ex approx resultExP E_y frack_Cr^ left p cosvartheta hatr_y - p_y right frack_Cr^ left p cosvartheta hatr_y - right EyF frack times pWleftrright^ times left times costh times sinth right Ey approx resultEyP E_z frack_Cr^ left p cosvartheta hatr_z - p_z right frack_Cr^ left p cosvartheta - right abcliste
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Electric Dipole by by
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Electric Field by by