Question

The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25 ✕ 105 V/m.What is the maximum energy that can be stored in the capacitor?

Answer 1

Step-by-step explanation:

The given data is as follows.

Dielectric constant, K = 3.0

Area of the plates (A) = 0.021
`m^(2)`

Distance between plates (d) =
`2.75 * 10^(-3) m`

Maximum electric field (E) =
`3.25 * 10^(5) V/m`

Now, we will calculate the capacitance as follows.

C =
`(k \epsilon_(o) * A)/(d)`

=
`(3.0 * 8.85 * 10^(-12) * 0.021)/(2.75 * 10^(-3))`

=
`(0.55755 * 10^(-12))/(2.75 * 10^(-3))`

=
`0.203 * 10^(-9)` F

Formula to calculate electric charge is as follows.

E =
`(\sigma)/(k \epsilon_(o))`

or, Q =
`E * k * \epsilon_(o)A` (as
`(\sigma)/(\epsilon_(o)) = (Q)/(A)`)

=
`3.25 * 10^(5) * 3.0 * 8.85 * 10^(-12) * 0.021`

=
`181.2 * 10^(-9) C`

Formula to calculate the energy is as follows.

U =
`(1 * Q^(2))/(2 * C)`

=
`((181.2 * 10^(-9) C)^(2))/(2 * 1.6691 * 10^(-9))`

=
`(32833.44 * 10^(-18))/(3.3382 * 10^(-9))`

=
`9835.67 * 10^(-9)`

or, =
`98.35 * 10^(-7) J`

Thus, we can conclude that the maximum energy that can be stored in the capacitor is
`98.35 * 10^(-7) J`.