Question

A paper-filled capacitor is charged to a potential difference of V 0 = 2.5 V V0=2.5 V . The dielectric constant of paper is κ = 3.7 κ=3.7 . The capacitor is then disconnected from the charging circuit and the paper filling is withdrawn, allowing air to fill the gap between the plates. Find the new potential difference V 1 V1 of the capacitor.

Answer 1

`V_1=9.25 V`

Step-by-step explanation:

We are given that

`V_0=2.5 V`

`k=3.7`

We have to find the new potential difference of the capacitor.

When the capacitor is disconnected then the charge stored in capacitor is constant.

When we introduce material of dielectric constant k between the plates of capacitor then the capacitance of capacitor increases k times.

`C_0=(Q)/(V_0)`

`C'=kC=(kQ)/(V_1)`

`(Q)/(V_0)=(kQ)/(V_1)`

`V_0=(V_1)/(k)`

Using the formula

`V_0=(V_1)/(k)`

`V_1=V_0k=2.5* 3.7=9.25 V`

Hence, the new potential difference
`V_1=9.25 V`