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 The capacitor is then disconnected from the charging circuit and the paper filling is withdrawn, allowing air to fill the gap between the plates.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

Answer:

$V_1=9.25 V$

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.

$C_0=\frac{Q}{V_0}$

$C'=kC=\frac{kQ}{V_1}$

$\frac{Q}{V_0}=\frac{kQ}{V_1}$

$V_0=\frac{V_1}{k}$

Using the formula

$V_0=\frac{V_1}{k}$

$V_1=V_0k=2.5\times 3.7=9.25 V$

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