You make a capacitor from 2 flat plates each with an area of 10 cm^2 you use a 1mm thick piece of pyrex glass as your dielectric. a. a.what is the maximum voltage you can safely apply to this capacitor?
b. what is its capacitance?
c. what is the maximum amount of charge you can store?
d. when this much charge is on the capacitor, how much energy is stored?

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aristocles aristocles · Answer 1 · 2019-07-17T03:11:01+02:00

Answer:

Part a)

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]E = 4.86 \times 10^{-3} J[/tex]

Explanation:

Part a)

As we know that dielectric constant of pyrex glass is 5.6 and its dielectric breakdown strength is given as

[tex]E = 14 \times 10^6 V/m[/tex]

now we have

[tex]E . d = \Delta V[/tex]

[tex](14 \times 10^6)(0.001) = \Delta V[/tex]

so we have

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

Capacitance is given as

[tex]C = \frac{k\epsilon_0 A}{d}[/tex]

[tex]C = \frac{5.6(8.85 \times 10^{-12})(10 \times 10^{-4}}{0.001}[/tex]

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

Now we have

[tex]Q = C\Delta V[/tex]

[tex]Q = (4.96 \times 10^{-11})(14 \times 10^3)[/tex]

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]Energy = \frac{1}{2}CV^2[/tex]

[tex]E = \frac{1}{2}(4.96 \times 10^{-11})(14 \times 10^3)^2[/tex]

[tex]E = 4.86 \times 10^{-3} J[/tex]