Two identical parallel plate air capacitors are connected in series to a battery of emf V. If one of the capacitor is completely filled with dielectric material of constant K, then potential difference of the other capacitor will become

Options

A. `(K)/(V(K+1))`
B. `(KV)/(K+1)`
C. `(K-1)/(KV)`
D. `(V)/(K(K+1))`

Answer

B

Explanation

When A is filled with a material of dielectric constant K,
its capacity becomes KC.
If `V_(A) and V_(B)` are the voltages across A and B, then
`V_(A)+V_(B)=V` . . (1)
The same charge (Q) flows through them and Q=CV
`therefore KCA_(A)=CV_(B)`
`therefore (V_(A))/(V_(B))=(C)/(KC)=(1)/(K)" "therefore V_(A)=(V_(B))/(K)`
`therefore` From (1), `(V_(B))/(K)+V_(B)=V`
`(V_(B)+KV_(B))/(K)=V" "therefore (K_(B)(K+1))/(K)=V`
`therefore` The P.D. across the other capacitor B is `V_(B)=(KV)/(K+1)`