Question

(a) What is the capacitance of a parallel plate capacitor having plates of area $1.50\textrm{ m}^2$ that are separated by 0.0200 mm of neoprene rubber? (b) What charge does it hold when 9.00 V is applied to it?

- $4.45\textrm{ }\mu\textrm{F}$
- $4.00\times 10^{-5}\textrm{ C}$

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This is College Physics Answers with Shaun Dychko. What is the capacitance of a parallel plate capacitor that has an area of 1.50 square meters, a separation between plates of 0.0200 millimeters and a material called 'neoprene' put between the plates, which has a dielectric constant of 6.7 when we look up in table [19.1]? So the capacitance of a parallel-plate capacitor is dielectric constant multiplied by the permittivity of free space times the area of the plates divided by the separation between the plates. So that's 6.7 times 8.85 times 10 to the minus 12 farads per meter times 1.50 square meters divided by 0.0200 times 10 to the minus 3 meters and that's 4.45 microfarads. And part (b) asks if 9.00 volts is applied across the capacitor, how much charge will be stored on it? So capacitance is the amount of charge divided by the voltage and we multiply both sides by

*V*to solve for*Q*. So*Q*is the capacitance times the voltage and here's the capacitance from part (a) multiplied by 9.00 volts which gives 4.00 times 10 to the minus 5 coulombs of charge.