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.

Respuesta :

Answer:

Explanation:

capacitance of each capacitor

Câ‚€= Qâ‚€ / Vâ‚€

Vâ‚€ = Qâ‚€ / Câ‚€

New total capacitance = Câ‚€ ( 1 + K )

Common potential

= total charge / total capacitance

= 2 Qâ‚€ / [ Câ‚€ ( 1 + K ) ]

2 Vâ‚€ / ( 1 + K )

b )

Common potential = 2 x Vâ‚€ / ( 1 + 7.8 )

= .227  V₀

charge on capacitor with dielectric

= .227  V₀ x 7.8 C₀

= 1.77 Vâ‚€Câ‚€

= 1.77 Qâ‚€

Ratio required = 1.77