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Two identical air-filled parallel-plate capacitors C1 and C2 are connected in series to a battery that has voltage V. The charge on each capacitor is Q0. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K>1 is inserted between the plates of capacitor C1, completely filling the space between them.

What is the total positive charge stored on the two capacitors?
Express your answer with the appropriate units.

User Poy
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2 Answers

4 votes

Final answer:

The total positive charge stored on two series-connected parallel-plate capacitors remains 2Q0 even after inserting a dielectric into one of them. Capacitors in series have the same charge despite changes in voltage distribution after the insertion of the dielectric.

Step-by-step explanation:

When two identical air-filled parallel-plate capacitors, C₁ and C₂, are connected in series to a battery with voltage V and each has an initial charge of Q0, inserting a dielectric with dielectric constant K into one of them (C₁) while keeping the battery connected does not change the total charge in the system. The total charge remains the same because the battery ensures that the voltage across the entire series combination stays constant, and therefore, the series combination must hold the same total charge Q0. However, what does change is the voltage distribution across the two capacitors and the individual stored charge. The capacitor with the dielectric (C₁) will end up with a reduced voltage across it compared to the capacitor without the dielectric (C₂).

The total positive charge stored on the two capacitors is 2Q0, considering both the positive sides of C₁ and C₂. The initial charge on each capacitor is Q0, and since the capacitors are in series, the charge on each one remains the same even after the dielectric is inserted into C₁.

User Shubhaw Kumar
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Answer:

Step-by-step explanation:

In the first case,

Q₀ = C₁ V, where C₁ is equivalent capacitance of two capacitor having capacitance C each.

C₁ = C / 2

Q₀ = CV / 2

CV = 2Q₀

Total charge = 2Q₀ = CV

In the second case ,

capacitance will be kC and C .

equivalent capacitance

C₂ = kC x C / ( kC + C)

= kC / ( 1+k)

Charge on each capacitor = V x C₂

= kCV / ( 1+k)

= 2kQ₀ / ( 1+k)

This will be the charge on each capacitor. So

total charge = 4kQ₀ / ( 1+k)

User ProbablePattern
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