-1
$\begingroup$

If a parallel plate capacitor is fully filled with a conducting slab and is connected to a battery then how does current flows in the circuit? As the electric field between the capacitor plates is zero then how does charge flows? Does the capacitance of the capacitor infinite? Now consider an isolated capacitor which is charged and then we introduce a conducting slab then the whole charge on the capacitor should flow from one plate to the other and the capacitance should be zero but in all textbooks, it is said that introducing a conducting slab wilenter image description herel increase capacitance

$\endgroup$
1
  • $\begingroup$ After you fully insert a conducting slab in between the capacitor plates your capacitor won't be a "capacitor" any more. $\endgroup$ Mar 29 '19 at 6:55
0
$\begingroup$

Introduction of a conducting slab will not increase the capacitance, as you introduce a conductor that doesn't shorts the plates then the system will act like two capacitors in series, and hence my capacitance will decrease. Two capacitors will be formed between (first plate and conductor ) and (conductor and second plate). This way capacitance decreases .

When conductor's length fits in the capacitor it'll short circuit it and hence now it'll behave like a simple conductor and capacitance will decrease.

Hope that helps:) , Thanks

$\endgroup$
3
  • $\begingroup$ If we connect the capacitor with the conducting slab to a battery then does any current flows through the circuit $\endgroup$ Mar 29 '19 at 9:42
  • $\begingroup$ Same battery or another battery? Can you please make it clear what you're trying to ask $\endgroup$
    – amanasci
    Mar 29 '19 at 9:44
  • $\begingroup$ Same battery and the capacitor is still filled with the conducting slab $\endgroup$ Mar 29 '19 at 9:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.