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Two identical capacitors are connected parallel. Initially they are charged to a potential V₀ and each acquired a charge Q₀. The battery is
Question
Two identical capacitors are connected parallel. Initially they are charged to a potential V₀ and each acquired a charge Q₀. The battery is then disconnected, and the gap between the plates of one capacitor is filled with a dielectric κ.(a) What is the new potential difference V across the capacitors. possible asnwers: V=(V₀)^2/[kQ₀+V₀), V=V₀/2k, V=V₀/2, V=kQ₀/V₀, V=2V₀/[k+1](b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
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Physics
4 years
2021-08-28T00:48:16+00:00
2021-08-28T00:48:16+00:00 1 Answers
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Answer:
ΔV ’= 2Δv / (1 + k)
, Q’= Q₀ 1,772
Explanation:
When the capacitors are connected in series the capacitance is added
C_eq = C₁ + C₂2
C_eq = 2 C₀
ΔV = Q₀ / C_eq
ΔV = Q₀ / 2 C₀
In this case, a dielectric is introduced to one of the capacitors, so its capacity changes.
C₂’= k C₂
Since the two capacitors have the same initial value let’s call Co
C_eq’= C₀ (1 + k)
The set charge is
C_eq ’= Q₀ / ΔV’
ΔV ’= Q₀ / Ceq’
ΔV ’= Q₀ / C₀ (1 + k)
The relationship between these voltages is
ΔV’/ ΔV = 2 / (1 + k)
ΔV ’= 2Δv / (1 + k)
The initial charge of the capacitor is
Q₀ = C₀ ΔV
The charge after introducing the dielectric is
Q’= k C₀ ΔV’
Q ’= k C₀ 2 ΔV / (1 + k)
Q’= Q₀ 2k / (1 + k)
Q’= Q₀ 2 7.8 /(1+7.8)
Q’= Q₀ 1,772