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*Question:*
A parallel plate capacitor has a plate area of
and a plate
separation of cm. A potential difference of V is applied
across the plates with no dielectric present. The battery is then disconnected,
and a piece of Bakelite () is inserted which fills the region between
the plates. What is the capacitance, the charge on the plates, and the
potential difference between the plates, before and after the dielectric
is inserted?

*Answer:* Before the dielectric is inserted,
the space between the plates is presumably filled
with air. Since the dielectric constant of air is virtually
indistinguishable from that of a vacuum, let us use the vacuum formula (108) to
calculate the initial capacitance . Thus,

After the dielectric is inserted, the capacitance increases by a factor ,
which in this case is , so the new capacitance is given by

Before the dielectric is inserted, the charge on the plates is
simply

After the dielectric is inserted, the charge is exactly the same, since the
capacitor is disconnected, and so the charge cannot leave the plates.
Hence,

The potential difference before the dielectric is inserted is given as
V. The potential difference after the dielectric is
inserted is simply

Note, of course, that .

** Next:** Example 6.3: Equivalent capacitance
** Up:** Capacitance
** Previous:** Example 6.1: Parallel plate
Richard Fitzpatrick
2007-07-14