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. (a) What is the new potential difference V across the capacitors. possible asnwers: V=(Vo)^2/[kQo+Vo), V=Vo/2k, V=Vo/2, V=kQo/Vo, V=2Vo/[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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Khoii Minh 1 year 2021-09-03T15:44:05+00:00 1 Answers 1 views 0

Answers ( )

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    2021-09-03T15:45:21+00:00

    Answer:

    Explanation:

    capacitance of each capacitor

    C₀= Q₀ / V₀

    V₀ = Q₀ / C₀

    New total capacitance = C₀ ( 1 + K )

    Common potential

    = total charge / total capacitance

    = 2 Q₀ / [ C₀ ( 1 + K ) ]

    2 V₀ / ( 1 + K )

    b )

    Common potential = 2 x V₀ / ( 1 + 7.8 )

    = .227  V₀

    charge on capacitor with dielectric

    = .227  V₀ x 7.8 C₀

    = 1.77 V₀C₀

    = 1.77 Q₀

    Ratio required = 1.77

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )