21 Replies
@Gyro Gearloose
@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.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.how do i find the final charge on capacitor
the solution directly uses the final capacitance of the system
but i use dq = dc.V where dc= A epsilon/d where d is distance between the plates
dq= AVepsilon/d and then integrate with d ranging from do to d1
q = AVepsilonln(d1/do)
Your dC consideration is wrong
C=Ae/x => dC= -Ae.dx/x^2
Now integrate
This is dimensionally incorrect
oh dC galat likha meine
this gives q = AeV/d1 - AeV/do
still feels wrong
are my limits wrong?
dx also take as -dx since dist will reduce
The other wave around
Actually doesn't matter
wait why u doing that
cuz u cant just get final charge directly using final state
intermediate condition nahi consider karte?
i might be wrong capacitor thoda weak hai
hm
idk i think final state ka hi C and V se Q nikaalenge
but mera bhi capcitor weak hi hai
why so?
at the equilibrium position C and V is fixed so wouldnt charge flow until charge increases to Q/C = V?
at do V=0 . then V is applied and do starts to decrease due to little little charge getting accumulated. eventually eqbm is at d1
@hardcoreisdead does this help?
I'm considering that the time taken for charge to be induced initially is negligible
ig thats the only explanation then
otherwise entire capacitors chapter would collapse lol
Isn't V variable here?
Sorry, I'm dumb 🤦♂️
If there is a source connected, then won't the source act to maintain the voltage?
Yeah, I forgot about that part.
+solved @Opt
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