Electric potential doubt
Doubt in Q27
how is electric heating related to electric potential?
This is from cengage chapter 3
Heating effect is in chapter 6
22 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I don't even know where to start
Maybe conservation of energy somehow?
I'm pretty sure it's a weird way of asking energy change
Oh I see
Which makes sense sort of
So is it related to self energy somehow then?
Because that's the only thing I can think of related to this
Also wouldn't this be assuming that all of the energy is converted to heat
q = it
H = i²rt = Vit = Vq
Where V is potential difference.
The last time I did this was in grade 10. I mean, heat is taught afterwards and cengage doesn't mix concepts like that. Shouldn't there be another way
Ok, conservation of energy i suppose.
How tho
Assuming 100% of energy is in heat. How
Original energy is going to be partially stored in the spherical capacitor system, and partially lost as heat
That's more apt.
Either all the Kinetic and potential is converted to heat (idk how this will be solved,not enough data imo) or maybe the charge expands the shell and the excess energy is converted to heat?
This might be solvable. Might have to derive the formula like self energy tho
I thought that too but then again, capacitors is next chapter so you might be able to solve that but there should be a way to solve using only potential
Well the capacitance is derived from potential, so technically this comes first
True
Welp, howd we go about this though
@Opt how would I work out the math for this
Can you give me maybe the first 1-2 steps?
Self energy of a shell is Kq²/R iirc
KQ^2/2R
Then, charge transfer will occur until potential of both surfaces is equal, so do q1/R1 = q2/R2 as well as charge conservation to figure out charge on each
Is it?
I don't remember the derivation.
dw=Vdq=Kqdq/R
Integrate from 0 to Q
Ah right.
Nvm
But yeah, that should work
Yea but
How do I apply the formula
Alr I got it. Thank you
+solved @Opt
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