Capacitor
If the parallel plane capacitor is charged and then battery is disconnected and the plated moved farther apart then why electrostatic energy of capacitor increases
27 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.Charge is conserved.
Capacitance decreases
Q²/2C increases
@Opt bro can you explain briefly
Once the battery is disconnected, there is no charge flow possible, right? So the charge gets conserved.
Now, the energy is given by q²/2C
And C = Aε/d, so U = q²d/2Aε right?
Keeping area and permittivity constant, (and charge is already constant), we see that U is directly proportional to d
@Opt ok bro thanks thoda smjh aaya
@Opt bro kuch baat krna h dm kro
Kya samajh nahi aaya?
@Opt ok now understands me u = 1/2 cv² se k rha tha
Haan, par V constant nahi hota hain yahaan
That's the important point
@Opt hnn
If there is a cell, it will maintain V by changing q.
If there isn't a cell, q will be maintained by changing V.
@Opt but it's a battery
There isn't a battery na?
Disconnected
@Opt first it's connected with battery and after this it is discon
Disconnected*
Plate is already charged
Yeah, that battery just gives an amount of charge q to the capacitor
And then q is conserved
Yaa after separation q is conserved
Ok, so what are you not understanding ?
How the electrostatic energy of capacitor increased e is inversely propotional to d hota h na to wo decrease Hoga an
No, it's directly proportional
See this
@Opt ok ab properly aa gya
Thanks bro
Close the thread na
@Opt mtlb???
Use +solved and then ping the person
+solved @Opt
Like that basically
Yeh lo bhai
@Devalok Varshney thank u bhai
+solved @Devalok Varshney @Opt \
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