Chapter 8, Problem 8.1CYU

Check Your Understanding The capacitance of a parallel-plate capacitor is 2.0 pF. If the area of each plate is 2.4 cm², what is the plate separation?

To determine

The distance between the plates.

Answer to Problem 8.1CYU

$0.001062\text{m}$

Explanation of Solution

Given:

Capacitors, $C=2.0\text{pF=2}\text{.0}\times {\text{10}}^{-12}\text{F}$

Area, $A=2.4{\text{cm}}^2=2.4\times {\text{10}}^{-4}{\text{ m}}^2$

Formula used:

The capacitance of the capacitor is given by,

$\begin{array}{l}C=\frac{{\epsilon }_{0}A}{d}\\ \text{Where},\\ C=\text{Capacitance}\\ A=\text{Area}\text{of plate}\\ d=\text{Distance between the plates}\end{array}$

The capacitance of the capacitor is given by,

$\begin{array}{l}C=\frac{{\epsilon }_{0}A}{d}\\ 2.0\times {10}^{-12}=\frac{8.85\times {10}^{-12}\times 2.4\times {10}^{-4}}{d}\\ d=\frac{8.85\times {10}^{-12}\times 2.4\times {10}^{-4}}{2.0\times {10}^{-12}}\\ d=0.001062\text{m}\end{array}$

Conclusion:

The distance between the plates is, $d=0.001062\text{m}$.