The plates of a parallel-plate capacitor each have an area of 0.1 m^2 and are separated by a 0.9 mm thick layer of glass. the capacitor is connected to a 14 v battery. (the dielectric constant for glass is 5.) (a) find the capacitance.
(b) find the charge stored.
(c) find the electric field between the plates.

a)

b)

c)

Explanation:

a) The capacitance of a parallel-plate capacitor with plates of area A separated a distance d with a dielectric of dielectric constant k is given by the formula:

where is the permittivity.

We use our values in S.I.:

b) The charge stored Q on a capacitor of capacitance C connected to a voltage V is given by the formula Q=CV.

Using our values:

c) The electric field E between plates separated a distance d where a voltage V is applied is obtained with the formula .

We use our values:

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explanation:

v=ir = 3x15 =45 v