Q19.5-53PE - College Physics (Urone)

Question

Find the capacitance of a parallel plate capacitor having plates of area \(5.00\;{m^2}\) that are separated by \(0.100\;mm\)of Teflon.

Step-by-Step Solution

Answer

The parallel plate capacitor's capacitance is \(0.92{\rm{ }}\mu F\).

Step 1: Defining capacitor

A capacitor is a device used to store electrical energy and works in an electric field. It is a passive electrical component with two terminals. Capacitance is the term used to describe the effect of a capacitor.

Step 2: Work of Capacitor and Information Provided

Any pair of conductors separated by an insulating substance is referred to as a capacitor. When the capacitor is charged, the two conductors have charges of equal magnitude \(Q\) and opposite sign, and the positively charged conductor's potential \(\Delta V\) with respect to the negatively charged conductor is proportional to \(Q\) The ratio of \(Q\) to \(\Delta V\) determines the capacitance \(C\).

\(C = \frac{Q}{{\Delta V}}\)

The parallel plate capacitor's capacitance is \(C = \kappa \frac{{{\varepsilon _0}A}}{d}\)

Where \(A\) is the area of each plate, \({\rm{d}}\)is the distance between them, and \({\varepsilon _0}\)is the vacuum permittivity 

\(\begin{array}{c}{\varepsilon _0} = 1/(4\pi k)\\ = 8.854 \times {10^{ - 12}}{\rm{ }}{C^2}/\left( {N \times {m^2}} \right)\end{array}\)

If the plates are separated by vacuum, \(\kappa  = 1\); otherwise, the dielectric constant of the dielectric is \(\kappa  > 1\). 

The plates of the capacitor have the following area: \(A = 5.00\;{m^2}\)

The capacitor plates are separated by the following distance:

\(\begin{array}{c}d = (0.100\;mm)\left( {\frac{{1\;m}}{{1000\;mm}}} \right)\\ = 0.100 \times {10^{ - 3}}\;m\end{array}\)

Teflon's dielectric constant is: \(\kappa  = 2.1\)

Step 3: Value of the parallel plate capacitor’s capacitance

Equation is used to calculate the parallel plate capacitor's capacitance.

\(C = \kappa \frac{{{\varepsilon _0}A}}{d}\)

Substitute the values of \(\kappa \),\({\varepsilon _0}\),\(A\) and \(d\) -

\(\begin{array}{c}C = (2.1)\left[ {8.854 \times {{10}^{ - 12}}{\rm{ }}{C^2}/\left( {N \times {m^2}} \right)} \right]\frac{{5.00\;{m^2}}}{{0.100 \times {{10}^{ - 3}}\;m}}\\ = 0.92 \times {10^{ - 6}}\;F\\ = \left( {0.92 \times {{10}^{ - 6}}\;F} \right)\left( {\frac{{{{10}^6}{\rm{ }}\mu F}}{{1\;F}}} \right)\\ = 0.92{\rm{ }}\mu F\end{array}\)

 

Therefore, capacitance value is obtained as \(0.92{\rm{ }}\mu F\).