Potential difference in field of point charge
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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 -
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Exercise:
Show that the potential difference between two pos in the field of a po charge Q is given by Delta V_AB k_C Q left fracr_B - fracr_A right where r_A and r_B are the distances of pos A and B from the po charge respectively.
Solution:
We ase that the pos A and B are on the same radial field line in the electric field of the po charge Q. The electric field as a function of the distance to the po charge is given by Er k_C fracQr^ The work done by the electric field on a test charge q moving from A and B is W_Ato B _A^B vecF_E textdvecr _r_A^r_B q Er textdr k_C q Q _r_A^r_B fracr^ textdr k_C q Q left -fracr right_r_A^r_B k_C q Q left fracr_A - fracr_B right The potential difference between pos A and B is Delta V_AB -fracW_Ato Bq -frack_C q Q left fracr_A - fracr_B rightq k_C Q left fracr_B - fracr_A right quad square The pos with the same distance to the po charge as B define an equipotential surface since they can all be reached from B through a circular arc with radius r_B. The work along a circular arc is zero since the force is perpicular to the displacement. As a consequence the derivation is valid for any two pos A and B in the field of the po charge.
Show that the potential difference between two pos in the field of a po charge Q is given by Delta V_AB k_C Q left fracr_B - fracr_A right where r_A and r_B are the distances of pos A and B from the po charge respectively.
Solution:
We ase that the pos A and B are on the same radial field line in the electric field of the po charge Q. The electric field as a function of the distance to the po charge is given by Er k_C fracQr^ The work done by the electric field on a test charge q moving from A and B is W_Ato B _A^B vecF_E textdvecr _r_A^r_B q Er textdr k_C q Q _r_A^r_B fracr^ textdr k_C q Q left -fracr right_r_A^r_B k_C q Q left fracr_A - fracr_B right The potential difference between pos A and B is Delta V_AB -fracW_Ato Bq -frack_C q Q left fracr_A - fracr_B rightq k_C Q left fracr_B - fracr_A right quad square The pos with the same distance to the po charge as B define an equipotential surface since they can all be reached from B through a circular arc with radius r_B. The work along a circular arc is zero since the force is perpicular to the displacement. As a consequence the derivation is valid for any two pos A and B in the field of the po charge.
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Exercise:
Show that the potential difference between two pos in the field of a po charge Q is given by Delta V_AB k_C Q left fracr_B - fracr_A right where r_A and r_B are the distances of pos A and B from the po charge respectively.
Solution:
We ase that the pos A and B are on the same radial field line in the electric field of the po charge Q. The electric field as a function of the distance to the po charge is given by Er k_C fracQr^ The work done by the electric field on a test charge q moving from A and B is W_Ato B _A^B vecF_E textdvecr _r_A^r_B q Er textdr k_C q Q _r_A^r_B fracr^ textdr k_C q Q left -fracr right_r_A^r_B k_C q Q left fracr_A - fracr_B right The potential difference between pos A and B is Delta V_AB -fracW_Ato Bq -frack_C q Q left fracr_A - fracr_B rightq k_C Q left fracr_B - fracr_A right quad square The pos with the same distance to the po charge as B define an equipotential surface since they can all be reached from B through a circular arc with radius r_B. The work along a circular arc is zero since the force is perpicular to the displacement. As a consequence the derivation is valid for any two pos A and B in the field of the po charge.
Show that the potential difference between two pos in the field of a po charge Q is given by Delta V_AB k_C Q left fracr_B - fracr_A right where r_A and r_B are the distances of pos A and B from the po charge respectively.
Solution:
We ase that the pos A and B are on the same radial field line in the electric field of the po charge Q. The electric field as a function of the distance to the po charge is given by Er k_C fracQr^ The work done by the electric field on a test charge q moving from A and B is W_Ato B _A^B vecF_E textdvecr _r_A^r_B q Er textdr k_C q Q _r_A^r_B fracr^ textdr k_C q Q left -fracr right_r_A^r_B k_C q Q left fracr_A - fracr_B right The potential difference between pos A and B is Delta V_AB -fracW_Ato Bq -frack_C q Q left fracr_A - fracr_B rightq k_C Q left fracr_B - fracr_A right quad square The pos with the same distance to the po charge as B define an equipotential surface since they can all be reached from B through a circular arc with radius r_B. The work along a circular arc is zero since the force is perpicular to the displacement. As a consequence the derivation is valid for any two pos A and B in the field of the po charge.
Contained in these collections
| Title | Matched on |
|---|---|
| Potential Difference in Field of Point Charge | tagstitle |
| Potential difference in field of charged wire | tagstitle |
| Potential difference in field of charged plate | tagstitle |
| Potential Difference in Field of Charged Wire | title |
| Van de Graaff Accelerator | tags |