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.

Answer 1

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