Question

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 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?

Answer 1

Answer:

5.3 nC

Explanation:

The initial charge stored on the capacitor is given by:

$Q=CV$ (1)

where

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

$V=100 V$ is the potential difference across the capacitor

Substituting numbers into the equation, we have

$Q=(25\cdot 10^{-12} F)(100 V)=25\cdot 10^{-10}C=2.5 nC$

When the Teflon slab is inserted between the plates, the capacitance of the capacitor is increased as follows:

$C'=kC$

where k=2.1 is the dielectric constant of the Teflon. Since the voltage V remains constant, this means that the new charge stored by the capacitor (1) will be

$Q'=C'V=kCV=kQ$

and so

$Q'=(2.1)(2.5 nC)=5.3 nC$