Question

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.

Answer 1

a)
`C=4.92*10^(-9)F`

b)
`Q=6.88*10^(-8)C`

c)
`E=15555.6V/m`

Step-by-step 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:

`C=(k\epsilon_0A)/(d)`

where
`\epsilon_0=8.85*10^-12F/m` is the permittivity.

We use our values in S.I.:

`C=(k\epsilon_0A)/(d)=((5)(8.85*10^-12F/m)(0.1m^2))/((0.0009m))=4.92*10^(-9)F`

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:

`Q=CV=(4.92*10^(-9)F)(14V)=6.88*10^(-8)C`

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

We use our values:

`E=(V)/(d)=(14V)/(0.0009m)=15555.6V/m`