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IB Physics: Equipotentials 

Chris Doner
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21 окт 2024

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Комментарии : 19   
@JessieAllen-r6m
@JessieAllen-r6m 7 месяцев назад
Very helpful lesson. Thank you Mr Doner.
@donerphysics
@donerphysics 6 месяцев назад
You are very welcome
@emir2750
@emir2750 4 года назад
this video gives me french fries and ranch sauce vibes, thanks
@donerphysics
@donerphysics 4 года назад
Any time
@manitverma4425
@manitverma4425 2 года назад
bro what?!
@rosahu8826
@rosahu8826 3 года назад
thank you so so much! my teacher only went over parallel plates i didnt even know the rest of this was in the syllabus... THANK YOUUUUU 🙏
@donerphysics
@donerphysics 3 года назад
Glad I could help!
@aniekanumoren6088
@aniekanumoren6088 6 лет назад
Watching this before me exam. Thanks so much
@manitverma4425
@manitverma4425 2 года назад
Hi Mr. Chris, at 15:21 i dont understand how you got the g = GM/r^2 equation. In the data booklet its Fg = GMm/r^2. thank you in advance!
@donerphysics
@donerphysics 2 года назад
Just use g=F/m
@manitverma4425
@manitverma4425 2 года назад
@@donerphysics oh yeah! makes total sense now! Thank you for replying!
@kallesipila1330
@kallesipila1330 5 лет назад
The question in 10:24 you said that the E field stays constant. However, as E=deltaV/delta r , given the potential difference (deltaV) is kept constant, doesn't the electric field strength have to vary with the distance between the plates? The question didn't really specify but I'm guessing nobody would assume that the battery voltage would just change
@donerphysics
@donerphysics 5 лет назад
E= delta V/delta r is really a derivative. It applies to very (infinitessimally) small regions, but it is based on the idea that the Electric field between parallel plates is constant. We effectively consider two points in space that are so close together that the voltage and distance between points does not have enough space to change.
@kallesipila1330
@kallesipila1330 5 лет назад
@@donerphysics Thanks! Love your videos
@manitverma4425
@manitverma4425 2 года назад
I had the same question! But after reading YOUR question I got what i did wrong!
@skodsrs9453
@skodsrs9453 4 года назад
At 9:48, if those electric potential values (10V and 20V) were negative, then the electric field lines would be from left to right (because -10 > -20 and because electric field lines ALWAYS move from higher to lower potential). Is this assessment correct?
@donerphysics
@donerphysics 4 года назад
yup
@franksinatra4736
@franksinatra4736 7 лет назад
i do not really get why the magnitude of the electric field exprience by a charge between to plates (equipotential lines) is the same no matter what is the position of the charge
@donerphysics
@donerphysics 6 лет назад
Because the E field between plates is uniform. Though the potential energy depends on position, the force field is uniform.
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