Question

An air-filled parallel-plate capacitor is constructed with a plate area of 0.40 m2 and a plate separation of 0.10 mm. It is then charged to a potential difference of 12 V? (ε0 = 8.85 × 10-12 C2 /N ∙ m2 ) (a) How much charge is stored on each of its plates? (b) If the capacitor is then filled with a glass dielectric (K = 5.0), by how much does the charge if it changes at all? The capacitor is not connected to a battery.

Answer 1

The charge stored is
`Q = 4.25 *10^(-7 ) \ C`

The energy stored is
`E = 2.55*10^(-6) \ J`

Step-by-step explanation:

From the question we are told that

The area of the plates is
`A = 0.40 \ m^2`

The separation between the plate is
`d = 0.10 \ mm =0.0001 \ m`

The potential difference is
`V = 12 \ V`

The permitivity of free space is
`\epsilon_o = 8.85 *10^(-12) C^2 \cdot N^(-1) \cdot m^2`

The dielectric constant of glass is K = 5.0

Generally the capacitance of this capacitor is

`C = (\epsilon_o * A)/(d)`

substituting values

`C = (8.85*10^(-12) * 0.40)/(0.0001)`

`C = 3.34 *10^(-8) \ C`

The charge stored is mathematically evaluated as

`Q = CV`

substituting values

`Q = (3.54*10^(-8) * 12)`

`Q = 4.25 *10^(-7 ) \ C`

The energy stored is

`E = 0.5 * CV^2`

substituting values

`E = 0.5 * (4.25 *10^(-7) * 12^2)`

`E = 2.55*10^(-6) \ J`