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

0.275 nF

Step-by-step 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`