Question

A certain dielectric with a dielectric constant = 24 can withstand an electric field of 4 107 V/m. Suppose we want to use this dielectric to construct a 0.89 µF capacitor that can withstand a potential difference of 1930 V. (a) What is the minimum plate separation? (b) What must the area of the plates be?

Answer 1

Step-by-step explanation:

Electric field E = 4 x 10⁷ V / m

Dielectric constant k = 24

capacitance of capacitor

C = kε₀ A / d

d = plate separation

A = plate area

C = .89 x 10⁻⁶

V / d = electric field

for minimum d , electric field will be maximum

V / d = 4 x 10⁷

1930 / d = 4 x 10⁷

d = 1930 / 4 x 10⁷

d = 482.5 x 10⁻⁷ m

= 48.25 x 10⁻⁶ m

C = kε₀ A / d

.89 x 10⁻⁶ = 24 ε₀ A / d

A = .89 x 10⁻⁶ X d / 24 ε₀

A = .89 x 10⁻⁶ X 48.25 x 10⁻⁶ / 24 x 8.85 x 10⁻¹²

= 42.9 / 212.4

= .2019 m²

Answer 2

a)
`d=0.4699\,m`

b)
`A=1969.1\,m^2`

Step-by-step explanation:

Given that:

dielectric constant,
`k=24`

electric field,
`E=4107 \,V.m^(-1)`

capacitance,
`C=0.89* 10^(-6)\,F`

potential difference,
`V=1930\,V`

(a)

We know for two parallel plates is given by:

`V=E.d`

where: d= distance between the plates.

`1930=4107* d`

`d=0.4699\,m`

(b)

For parallel plate capacitor we have:

`C=(k.\epsilon_0.A)/(d)`

where:

A= area of plates

permittivity of free space,
`\epsilon_0=8.85* 10^(-12)\,F.m^(-1)`

`0.89* 10^(-6)=(24* 8.85* 10^(-12)* A)/(0.4699)`

`A=1969.1\,m^2`