Question

A 100 pF parallel-plate capacitor is made of two metal plates. The area of each plate is 100 mm2. What is the separation between the plates?Group of answer choices9.06x10-6 m8.85x10-6 m6.44x10-6 m5.60x10-6 m1.13x10-6 m

Answer 1

The capacitance C of a parallel-plate capacitor whose plates have an area A and are separated by a distance d is:

$C=\frac{\varepsilon_0A}{d}$

Where ε₀ is the vacuum permitivity:

$\varepsilon_0=8.854\times10^{-12}\frac{F}{m}$

Notice that the capacitance of the parallel-plate capacitor is given, as well as the area of its plates. Then, isolate d from the equation:

$\Rightarrow d=\frac{\varepsilon_0A}{C}$

Replace A=100mm^2, C=100pF and the value of vacuum permitivity to find the separation between the plates:

$d=\frac{(8.854\times10^{-12}\frac{F}{m})(100\times10^{-6}m^2)}{(100\times10^{-12}F)}=8.854\times10^{-6}m$

Therefore, the correct choice is: 8.85x10^-6m.