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

Answer:

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

Answer:

a) $d=0.4699\,m$

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

Explanation:

Given that:

dielectric constant, $k=24$

electric field, $E=4107 \,V.m^{-1}$

capacitance, $C=0.89\times 10^{-6}\,F$

potential difference, $V=1930\,V$

(a)

We know for two parallel plates is given by:

$V=E.d$

where: d= distance between the plates.

$1930=4107\times d$

$d=0.4699\,m$

(b)

For parallel plate capacitor we have:

$C=\frac{k.\epsilon_0.A}{d}$

where:

A= area of plates

permittivity of free space, $\epsilon_0=8.85\times 10^{-12}\,F.m^{-1}$

$0.89\times 10^{-6}=\frac{24\times 8.85\times 10^{-12}\times A}{0.4699}$

$A=1969.1\,m^2$