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

According the question, in first case (i.e. without dielectric)

For 2nd case when dielectric is inserted in one of the capacitor,

Given `V_(1)+V_(2)=V`
The capacitance of first capacitor becomes KC, So by charge conservation,
`(KC)V_(1)=CV_(2)" ...(ii)"`
`(V_(1))/(V_(2))=(1)/(K)`
`rArr" "(V_(1)+V_(2))/(V_(2))=(1+K)/(K)`
`rArr" "V_(2)=((KV)/(1+K))" "(because V_(1)+V_(2)=V)`