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What is the capacitance of a parallel plate capacitor with plates of area 1.50 m^2 separated by 0.0200 mm of neoprene rubber?

a) 1.77 × 10^(-10) F
b) 3.55 × 10^(-10) F
c) 7.07 × 10^(-11) F
d) 1.42 × 10^(-10) F
What charge does it hold when 9.00 V is applied to it?

a) 1.20 × 10^(-8) C
b) 2.50 × 10^(-8) C
c) 4.04 × 10^(-8) C
d) 6.36 × 10^(-9) C

1 Answer

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Final answer:

The capacitance of a parallel plate capacitor with the given specifications is approximately 3.55 x 10^-10 F. When a voltage of 9.00 V is applied to it, the charge held by the capacitor is approximately 3.20 x 10^-9 C.

Step-by-step explanation:

To calculate the capacitance of a parallel plate capacitor, we use the formula:


C = ε0 εr A / d


where C is the capacitance, ε0 is the vacuum permittivity (8.85 x 10-12 F/m), εr is the relative permittivity of neoprene rubber (typically about 6.7), A is the area of the plates (1.50 m2), and d is the separation between the plates (0.0200 mm or 0.0200 x 10-3 m).


Plugging in the values, we get:


C = (8.85 x 10-12 F/m) * 6.7 * (1.50 m2) / (0.0200 x 10-3 m)


This results in a capacitance of approximately 3.55 x 10-10 F.

For the charge held by the capacitor when a voltage of 9.00 V is applied, we use the formula:


Q = C * V


where Q is the charge, and V is the voltage applied.

Q = (3.55 x 10-10 F) * (9.00 V)


The charge it holds is thus approximately 3.20 x 10-9 C.

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