Question

A parallel plate capacitor has an area of A=200 cm^-2    and 0.50 cm plate distance. The capacitor is firstly connected to a power supply that has potential difference V0=30V    and after charged the power supply is disconnected. After then a sheet of insulating plastic material is inserted between the plates (completely filling the space between them). With the plastic material the potential difference decreases to 10 V while the charge on capacitor remains the constant. Find ;
a) the original (while filled air) capacitance C0 .

b) the stored charge Q

c) the capacitance C after the dielectric is inserted.

d) the dielectric constant of the dielectric.

e) the original electric field E0 between the plates.

f) The stored energy at the original capacitor.

Answer 1

PART A:

The original capacitance is (in SI units)

So, the answer to this part is 35.416 pF.

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PART B:

The stored charge is given by

So, the answer to this part is 1.06248 nC.

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PART C:

After the dielectric is inserted the potential difference reduces to 10 V. So, the new capacitance is

So, the answer to this part is 106.248 pF.

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PART D:

The capacitance is given by

So, the dielectric constant of the material is 3.

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PART E:

The original electric field was

So, the original electric field was 6000 V/m.

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PART F:

The stored energy in the original capacitor was

So, the stored energy was 15.9372 nJ.