Question about charged capacitors and inserting a dielectric into one

In summary, when a capacitor is connected to a battery, it accumulates charges until the potential difference equals that of the battery. When a dielectric is added between two capacitors, the potential difference across it decreases and one capacitor starts acting like a battery, providing charge to the other. This provision process ends when the potential difference between the two capacitors becomes equal. During this process, the total charge is conserved and can be determined using a set of equations.
  • #1
Kaushik
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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?
 
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  • #2
Kaushik said:
providing charge to the capacitor 1
And when does this 'provision' come to an end ?
 
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  • #3
BvU said:
And when does this 'provision' come to an end ?
When the potential difference between capacitor 1 and capacitor 2 become equal?
 
  • #4
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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Related to Question about charged capacitors and inserting a dielectric into one

1. What is a charged capacitor?

A charged capacitor is an electronic component that stores electrical energy in the form of an electric charge. It consists of two conductive plates separated by an insulating material, known as a dielectric.

2. How does a capacitor store charge?

A capacitor stores charge by creating an electric field between its two plates. When a voltage is applied, electrons from one plate are attracted to the other plate, creating a buildup of charge on both plates.

3. What happens when a dielectric is inserted into a charged capacitor?

When a dielectric is inserted into a charged capacitor, it increases the capacitance of the capacitor. This is because the dielectric material has a higher permittivity, which allows for more charge to be stored on the plates.

4. How does a dielectric affect the electric field in a capacitor?

A dielectric material reduces the strength of the electric field in a capacitor. This is because the electric field lines are shortened due to the presence of the dielectric, which decreases the voltage across the plates.

5. Can a dielectric be inserted into any type of capacitor?

Yes, a dielectric can be inserted into any type of capacitor. The effect of the dielectric on the capacitance will depend on the type of capacitor and the properties of the dielectric material being used.

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