Question

A 25 {\rm pF} parallel-plate capacitor with an air gap between the plates is connected to a 100 {\rm V} battery. A Teflon slab is then inserted between the plates, and completely fills the gap.What is the change in the charge on the positive plate when the Teflon is inserted? (change in q=? nC)

Answer 1

Answer:

0.275 nF

Explanation:

First of all, let's calculate the initial charge stored by the capactiro when it is in air. This is given by:

$Q_0 = C_0 V$

where

$C_0 = 25 pF=25\cdot 10^{-12} F$ is the initial capacitance of the capacitor

$V=100 V$ is the voltage of the battery

Substituting, we find

$Q_0 = (25\cdot 10^{-12}F)(100 V)=25\cdot 10^{-10} F=0.25 nF$

When the Teflon slab is inserted between the plates, the capacitance changes according to:

$C'=kC_0$

where

$k=2.1$ is the dielectric constant of Teflon. Therefore, the new charge stored on the capacitor will be:

$Q'=C' V=(kC_0)V=(2.1)(25\cdot 10^{-12} F)(100 V)=0.525 nF$

And so, the change in the charge on the capacitor is:

$\Delta Q=Q'-Q=0.525 nF-0.25 nF=0.275 nF$