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))`

Explanation

According to 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`
`(V_1)/V_2=1/K`
`Rightarrow (V_1+V_2)/V_2=(1+K)/K`