Two parallel plate air capacitance of same capacity C are connected in series to a battery of emf E. Then one of the capacitors is completely filled with dielectric material of constant K. The change in the effective capacity of the series combination is

Options

A. `(C)/(2)[(K-1)/(K+1)]`
B. `(2)/(C)[(K-1)/(K+1)]`
C. `(C)/(2)[(K+1)/(K-1)]`
D. `(C)/(2)[(K-1)/(K+1)]^(2)`

Answer

A

Explanation

Two parallel plate capacitors are connected in series,
`therefore` Theire resultant capacity `C_(1)=(CxxC)/(C+C)=(C)/(2)`
and when one of then is completely filled with a dielectric, its capacity becomes KC.
`therefore` new resultant capacity
`(1)/(C_(2))=(1)/(C)+(1)/(KC)=(KC+C)/(C*(KC))=(C(K+1))/(C(KC))`
`therefore C_(2)=(KC)/(K+1)`
`therefore` The change in the capacitance
`DeltaC=C_(2)-C_(1)`
`=(CK)/(K+1)=(C)/(2)`
`=C[(K)/(K+1)-(1)/(2)]`
`=C[(2K-K-1)/(2(K+1))]`
`DeltaC=(C)/(2)[(K-1)/(K+1)]`