Question

Two $4.0cm\times 4.0cm$metal plates are separated by a $0.20mm$-thick piece of Teflon.

a. What is the capacitance?

b. What is the maximum potential difference between the plates?

Short answer

a. The capacitance is$0.15nF$,

b. The maximum potential difference between the plates is$12kV$.

Step-by-step solution

Part (a)Step 1: Given information

We need to find the capacitance.

Part (a) Step 2: Simplify 

a. The capacitance will be given by

$C=\in {\in }_0\frac{A}{d}=\in {\in }_0\frac{L^2}{d}$,

Where $L$is the side of the square capacitor plates and $d$is the thickness of the teflon with dielectric constant $\in$(sometimes noted as $k$).

This means that we can find the value of the capacitance of our capacitor as

localid="1648665207896" $C=(2.1)(8.85\times {10}^{-12})\left(\frac{0.{04}^2}{2\times {10}^{-4}}\right)=1.5\times {10}^{-10}F$.

Part (b) Step 1: Given Information

We need to find the maximum potential difference between the plates.

Part (b) Step 2: Simplify

We are given that the maximum electric field strength that Teflon can withstand is $E_{max}=6\times {10}^{6}V/m$. Knowing that, the maximum potential difference through distance $d$can be found as

$∆V_{max}=E_{max}d$

Numerically, we will have

localid="1648665350833" $∆V_{max}=(6\times {10}^7)(2\times {10}^{4}V)$

*Both values, for the dielectric constant and the breakdown electric field were obtained from table $26.1$in the book.