Part A
A parallel-plate capacitor is made from two aluminum-foil sheets, each 6.4 cm wide and 5.7 m long. Between the sheets is a Teflon strip of the same width and length that is 2.8\times 10^{-2} mm thick.
For the steps and strategies involved in solving a similar problem, you may view the following Example 20-14 video:
What is the capacitance of this capacitor? (The dielectric constant of Teflon is 2.1.)
Express your answer in microfarads.
SOLUTION
First, determine the capacitance:
$C = \frac{\kappa A}{d} = \frac{(8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2})(0.0280 m^2)}{0.550 \times 10^{-3} m} = 4.505 \times 10^{-10} F$
C = 2.4
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\mu F
