capacitor
why cant we just add add 1/2CV^2 and 1/2(3C)V^2...also what do they mean by equivalent capacitance since no battery??
13 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.a battery causes potential difference to exist.
if there is no battery, there is no pot diff
there is a potential at steady state which they are saying will be total charge/total capacity
why will you add the two? can you walk us through the logic of that?
Potential diff can exist if u have a charged and uncharged capacitor connected through a wire
Ig in that case u can call the capacitor a battery
Make sure to send the whole question next time ok? Just advice.
I'm guessing the situation is that capacitor C is charged across a cell of EMF ε, then connected across an empty capacitor 3C.
Also, you can do 0.5CV²+1.5CV² here, there's no issue since the capacitors are in parallel. If they were in series however, you would need to consider the potential differences across each, so keep that in mind.
Also, equivalent capacitance can be deduced from how the voltage develops across the capacitor.
The upper part is positively charged and the lower is negatively charged so the entire system behaves like a single capacitor with (1+3)C capacitance
interesting. yea sure that is the 'before' state. the 'after' state would be no pd right?
yepp
But then U2 is coming 10/16 CE^2
Also how would you connect 2 capacitors in series without a battery....charge will flow till both capacitors are at equal potential ergo they're in parallel
Exactly
2CV² = 2C(ε/4)² which what they've gotten as well....
But the other capacitors is 3C
Oh wait
Exactly. That’s what’s happening
+solved @Opt
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