A25 {\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)

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:

where

is the initial capacitance of the capacitor

is the voltage of the battery

Substituting, we find

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

where

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

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

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