Question

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is 1.20×106 volts per meter.

(A) Compute the magnitude of the charge per unit area sigma on the conducting plate.
(B) Compute the magnitude of the charge per unit area sigma1 on the surfaces of the dielectric.
(C) Find the total electric-field energy U stored in the capacitor.

Answer 1

A) σ = EKε₀

σ – magnitude of a charge per unit area (conducting plate)

E – resultant electric field

K – 3.6 [dielectric constant]

ϵ₀ – 8.85 × 10⁻¹² F∙m⁻¹ [permittivity of free space]

σ =
`(1.20*10^(6))(3.6)(8.85*10^(-12))`

σ = 3.82×10⁻⁵ C/m²

B) σ₁ = σ(1 - 1 / K)

σ₁ – magnitude of the charge per unit area (surfaces of the dielectric)

σ₁ =
`3.82*10^(-5) (1-(1)/(3.6) )`

σ₁ = 2.76×10⁻⁵ C/m²

C) U = (AdKE² ε₀) / 2

U – total electric-field energy

A – area

d – diameter

U =
`((2.5*10^(-4)m^(2))(1.8*10^(-3)m)(3.6)(1.20*10^(6))^(2)(8.85*10^(-12)))/(2)`

U = 1.03×10⁻⁵ J

Answer 2

A). σ = 3.823 x
`10^(-5)`
`C^(2)`/N-
`m^(2)`

B).
`\sigma ^(')=2.76* 10^(-5)` C/
`m^(2)`

C).
`U=10.322` J

Step-by-step explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

`\sigma =k.\varepsilon _(0).E`

where, E is resultant electric field = 1.2 x
`10^(6)` V/m

`\varepsilon _(0)` is permittivity of free space = 8.85 x
`10^(-12)`
`C^(2)`/N-
`m^(2)`

k is dielectric constant = 3.6

∴
`\sigma =k.\varepsilon _(0).E`

= 3.6 x 8.85 x
`10^(-12)` x 1.2 x
`10^(6)`

= 3.823 x
`10^(-5)`
`C^(2)`/N-
`m^(2)`

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

`\sigma ^(')=\sigma\left ( 1-(1)/(k) \right )`

`\sigma ^(')=3.823* 10^(-5)\left ( 1-(1)/(3.6) \right )`

`\sigma ^(')=2.76* 10^(-5)` C/
`m^(2)`

C).

Area of the plate, A = 2.5
`cm^(2)`

= 2.5 x
`10^(-4)`
`m^(2)`

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

`U=(1)/(2)k\varepsilon _(0)E^(2)Ad`

`U=(1)/(2)* 3.6*8.85 *10^(-12)*\left ( 1.2* 10^(6) \right ) ^(2)* 2.5* 10^(-4)* 1800`

`U=10.322` J