How much energy is stored in the electric field of a 50-μm-diameter cell with a 7.0-nm-thick cell wall whose dielectric constant

Question

zhuklara [117]

2 years ago

15

How much energy is stored in the electric field of a 50-μm-diameter cell with a 7.0-nm-thick cell wall whose dielectric constant

is 9.0?

Physics

2 answers:

Nady [450] · Answer 1 · 2022-08-13T20:02:38+03:00

:<span> </span><span>Under the assumption that a cell is made up of two concentric spheres you find the surface are of the inside sphere which will be your A.

You already have your separation and dielectric constant so just use the formula you stated towards the end of your question and you get 8.93x10^-11 Farads which is about 89pF</span>

hichkok12 [17] · Answer 2 · 2022-08-13T22:04:08+03:00

The capacitance of the capacitor will be $\boxed{8.9\times10^{-11}\text{ F}}$ or $\boxed{89\text{ pF}}$ .

Explanation:

Given:

The diameter of the plates of the capacitor is $50\,\mu\text{m}$ .

The distance between the plates is $7\text{ nm}$ .

The dielectric constant of the medium is $9.0$ .

Concept:

The two conducting plates kept parallel to each other and placed at a particular distance from one another are considered to be the parallel plate capacitor.

The capacitance of the parallel plates capacitor formed by the two plates is given by the expression:

$\boxed{C=\dfrac{K\epsilon_0A}{d}}$

Here, $C$ is the capacitance of the plates, $K$ is the dielectric constant of the medium, $\epsilon_0$ is the permittivity of free space, $A$ is the area of the plates and $d$ is the distance between the plates.

The area of the plates is:

$\begin{aligned}A&=\pi R^2\\&=3.14\times(25\times10^{-6})^2\\&=1.96\times10^{-9}\text{ m}^2\end{aligned}$

Substitute the values in equation of capacitance.

$\begin{aligned}C&=\dfrac{9.0\times8.85\times10^{-12}\times1.96\times10^{-9}}{7.0\times10^{-9}}\\&=8.9\times10^{-11}\text{ F}\\&=89\text{ pF}\end{aligned}$