Question about charged capacitors and inserting a dielectric into one

  • #1
Kaushik
282
17
Homework Statement:
The parallel combination of two air filled parallel plate capacitors of capacitance ##C## and ##nC## is connected to a battery of voltage, ##V##. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant ##K## is placed between the two plates of the first capacitor. The new potential difference of the combined system is?
Relevant Equations:
##Q = CV##
##C_{eq}= C_1 + C_2## for parallel
## \frac{1}{C_{eq}}= \frac{1}{C_1} + \frac{1}{C_2}## for series
First when it is connected to the battery, the capacitors start accumulating charges such that the potential difference equals that of the battery. Then the current stops flowing.
##Q_1 = CV##
##Q_2 = nCV##
Where 1 and 2 represent the capacitor with capacitance C and nC respectively

Then, when we remove the battery and add a dielectric between the capacitor of capacitance ##C## (##C_{new} = KC##), the potential difference across it reduces. Now, the capacitor 2 starts behaving like a battery, providing charge to the capacitor 1.
After this, how can I proceed?
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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providing charge to the capacitor 1
And when does this 'provision' come to an end ?
 
  • #3
Kaushik
282
17
And when does this 'provision' come to an end ?
When the potential difference between capacitor 1 and capacitor 2 become equal?
 
  • #4
BvU
Science Advisor
Homework Helper
15,269
4,245
Correct.
And what is conserved during this provision process ?
Set up a set of equations before and after to find the new ##V##.
 

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