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

yanalaym

1 month 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:

Keith_Richards [3.2K] · Answer 1 · 2026-04-10T11:37:26+03:00

:<span> </span><span>Assuming a cell comprises two concentric spheres, you would calculate the surface area of the inner sphere, which will be your A.

Having already determined your separation and dielectric constant, simply apply the formula mentioned near the end of your query and calculate 8.93x10^-11 Farads, approximately 89pF.</span>

kicyunya [3.2K] · Answer 2 · 2026-04-10T11:35:50+03:00

The capacitor's capacitance will amount to $\boxed{8.9\times10^{-11}\text{ F}}$ or $\boxed{89\text{ pF}}$ .

Explanation:

The parameters given include:

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

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

The dielectric constant of the material is $9.0$ .

Concept:

When two conducting plates are positioned parallel to each other and spaced apart, they form a parallel plate capacitor.

The capacitance for the capacitor created by these two plates can be determined by the following equation:

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

In this equation, $C$ represents the capacitance, $K$ is the medium's dielectric constant, $\epsilon_0$ denotes the permittivity of free space, $A$ signifies the area of the plates, and $d$ indicates the distance between the plates.

The area of the plates is:

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

Substituting the values into the capacitance formula will yield your answer.

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