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The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise identical capacitor with an air gap, and it increases the maximum potential difference the capacitor can support. If the electric field in a material is sufficiently strong, the material will suddenly become able to conduct, creating a spark. The critical field strength, at which breakdown occurs, is 3.0 MV/m for air, but 60 MV/m for Teflon.
Part A
A parallel-plate capacitor consists of two square plates 16cm on a side, spaced 0.45mm apart with only air between them. What is the maximum energy that can be stored by the capacitor?
Part B
What is the maximum energy that can be stored if the plates are separated by a 0.45-mm-thick Teflon sheet?

Respuesta :

umohduke14

Answer: 580 x 10^-3 J

Explanation:

0.6mm is 0.6/1000 = 600*10^-6 m

The plate area is .17*.17 = 28.9*10^-3 m^2

Air:

The voltage that can be sustained by 0.60 mm of air dielectric is:

V = 3.0*10^6* 600*10^-6 = 1800 V

The capacitance is:

C = ε*A/d = 8.854*10^-12 * 28.9*10^-3/600*10^-6 = 426*10^-12 F = 426 pF

The energy stored in a capacitor is:

E = (1/2)*C*V^2 = (1/2)*426*10^-12*(1800)^2 = 691*10^-6 J

Teflon:

The voltage is:

V = 60*10^6* 600*10^-6 = 36*10^3 = 36 kV

According to the listed reference, the relative dielectric constant for teflon is 2.1, this figure multiplies the "ε" of free space.

The capacitance is:

C = ε*A/d = 2.1*8.854*10^-12 * 28.9*10^-3/600*10^-6 = 896*10^-12 F = 896 pF

It would have been easier to note that the capacitance is 2.1 times the air-dielectric case.

The maximum energy stored is:

E = (1/2)*C*V^2 = (1/2)* 896*10^-12* (36*10^3)^2 = 580*10^-3 J

The maximum energy that can be stored by parallel plate capacitor is 4.588×10⁻³ J and by taflon plate capacitor is 4.078 J.

What is the energy stored by a capacitor?

Energy stored by a capacitor is the electrical potential energy. The energy stored by a capacitor can be the given as,

[tex]E=\dfrac{1}{2}A\varepsilon_0kE_0^2d[/tex]

  • Part A-The maximum energy that can be stored by the capacitor-

For part A parallel-plate capacitor consists of two square plates 16 cm on a side, spaced 0.45 mm apart with only air between them.

Area of the plate with side 16 cm (0.16 m) is,

[tex]A=0.16\times0.16\\A=0.256\rm m^2[/tex]

The value of k is 1 for air. Thus, put the value in the formula as,

[tex]E=\dfrac{1}{2}(0.256)(8.85\times10^{-12})(1)(3.0\times10^{6})^2(0.45\times10^{-3})\\E=4.588\times10^{-3}\rm J\\[/tex]

  • Part B-The maximum energy that can be stored-

For part B plates separated by a 0.45-mm-thick Teflon sheet.

The value of k for Teflon is,

[tex]k=\dfrac{60}{3}\\k=20[/tex]

Thus, out the values again in the above formula as,

[tex]E=20(4.588\times10^{-3})\dfrac{20}{3}^2\\E=4.078\rm J[/tex]

Hence, the maximum energy that can be stored by parallel plate capacitor is 4.588×10⁻³ J and by taflon plate capacitor is 4.078 J.

Learn more about the energy stored by a capacitor here;

https://brainly.com/question/13578522

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