Question

(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.


pF

(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV

Answer 1

Explanation:

(a) Given that,

Area of a parallel plate capacitor, $A=1.8\ cm^2=1.8\times 10^{-4}\ m^2$

The separation between the plates of a capacitor, $d=0.01\ mm = 10^{-5}\ m$

The dielectric constant of, k = 2.1

When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :

$C=\dfrac{k\epsilon_o A}{d}$

Putting all the values we get :

$C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF$

(b) We know that the Teflon has dielectric strength of 60 MV/m, $E=60\times 10^6\ V/m$

The voltage difference between the plates at this critical voltage is given by :

$V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V$

or

V = 0.6 kV

Answer 2

We have that the Capacitance and potential difference is mathematically given as

• $Vmax=\frac{Q}{334.68pF}$
• C=334.68pF



Capacitance &potential difference

Question Parameters:

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.

a)

Generally the equation for the Capacitance  is mathematically given as

$C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}$

C=334.68pF

b)

Generally the equation for the Capacitance  is mathematically given as

$Vmax=\frac{Q}{C}$

Where

Q is the charge on the plates, and hence not given

Therefore, maximum potential difference is

$Vmax=\frac{Q}{334.68pF}$

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