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QCE Physics Questions Thread
Bri MT:
--- Quote from: jasmine24 on November 05, 2020, 07:15:33 am ---Hi, would anybody have any examples on "interpreting graphical representations of electrical potential difference vs electric current data to find resistance using its gradient and its uncertainty". I couldn't find any that also found the uncertainty and I'm unsure as to how to do that
Thank u!
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Hey,
I'll attach some images later today but remember that this just comes from Ohm's law V=IR => V/I = rise/run = gradient = R
In regards to uncertainty, what's the steepest line of best fit given your uncertainty? What's the least steep line of best fit? Constructing these will give you your range of values for R and hence your uncertainty.
Hope this helps!
K.Smithy:
Hi everyone!
I have been talking to a few people about this question and we haven't been able to come to an agreed upon answer :P
For part a, some think that the charge of Q2 is negative and Q1 is positive: due to the fact that they opposite charges the electric field strength would cancel out in the middle. Furthermore, the electric field strength is negative as you approach Q2 - therefore, Q2 has a negative charge.
Others think that both Q1 and Q2 are positive: Coulomb's law states that opposite charges attract and like charges repel - so that would explain the fact that the electric field strength is 0 in between the two point charges (if both are positive, they will repel each other).
I can see both sides, but I don't know which one is correct.
Thanks,
Katelyn :)
Bri MT:
Hi K.Smithy! :)
Q2 is positive, because of the zero as well as because it goes from positive to negative. Electric field strength is described in terms of the impact of a positive charge at that point in space, not in terms of how the two charges present would interact so the second pov doesn't really work despite it showing physics knowledge.
Good question - it's an interesting one that I can easily see being one people would argue about after an exam
K.Smithy:
--- Quote from: Bri MT on November 08, 2020, 01:28:08 pm ---Hi K.Smithy! :)
Q2 is positive, because of the zero as well as because it goes from positive to negative. Electric field strength is described in terms of the impact of a positive charge at that point in space, not in terms of how the two charges present would interact so the second pov doesn't really work despite it showing physics knowledge.
Good question - it's an interesting one that I can easily see being one people would argue about after an exam
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Awesome! Thank you so much :)
Bri MT:
--- Quote from: K.Smithy on November 08, 2020, 01:34:13 pm ---Awesome! Thank you so much :)
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No worries!
For questions like this I would really recommend drawing a field diagram to help you understand why the answer is what it is. Once you've drawn it you should be able to see that for positive -> negative the field is constantly from q1 to q2 but that for positive -> positive it isn't since both positives are trying to push positive charges away & therefore are acting in opposite directions.
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