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.

Question

Physics

Krish Gaines

25 November, 11:14

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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Vaughan · Answer 1

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