Our demonstration capacitor used in lab has circular plates 20 cm in diameter. L

Question

Our demonstration capacitor used in lab has circular plates 20 cm in diameter. Let us put the plates 5 mm apart. Note: the (relative) dielectric constant of Air is 1.00054 and the "dielectric strength" of Air is 3 MegaVolts/meter.

Part A - Voltage Rating

Assuming air between the plates, what is the maximum voltage this capacitor could hold before breakdown?

Give answer in volts.

What then is the maximum possible charge this capacitor could hold before breakdown?

Give answer in nC (nanoCoulombs)

Now let us fill the space between the plates with a dielectric (Strontium Titanate).  The dielectric constant of Strontium titanate is 310 and its dielectric strength is 8 MegaVolts/meter.  Now what is the maximum voltage the capacitor can have before breakdown?

Give answer in volts.

With the dielectric inserted, what is the maximum charge that the capacitor can hold (before breakdown)?

Give answer in nC (nanoCoulombs), so that you can easily compare with the answer in part C.

Explanation / Answer

Part A)

Apply V = Ed

V = (3 X 10^6)(.005)

V = 15000 V (1.50 X 10^4 V)

Part B)
Q = CV and

C = kEA/d

C = (1.00054)(8.85 X 10^-12)(pi)(.1)^2/.005

C = 5.56 X 10^-11 F

Now

Q = (5.56 X 10^-11)(15000)

Q = 8.35 X 10^-7 C

That is 835 nC

Part C)

V = Ed

V = (8 X 10^6)(.005)

V = 40000 V (4.0 X 10^4 V)

Part D)
Q = CV and

C = kEA/d

C = (310)(8.85 X 10^-12)(pi)(.1)^2/.005

C = 1.72 X 10^-8 F

Now

Q = (1.72 X 10^-8)(40000)

Q = 6.90 X 10^-4 C

That is 690000 nC (6.90 X 10^5 nC)

To find how much greater that is, divide 690000/835

That is 826 times more charge

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