Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is {0.50 10-3 C/m2, the cell wall is 5.0 nm thick, and the cell-wall material is air. (a) Find the magnitude of $\overrightarrow{\mathbf{E}}$in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10-16 m3. Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

The cells carrying charges resemble a parallel plate capacitor.

For the medium air, the expression of electric field is,

$\begin{array}{r}{E}_0=\frac{\sigma }{{\epsilon }_0}\\ =\frac{0.50\times {10}^{-3}\mathrm{C}/{\mathrm{m}}^2}{8.854\times {10}^{-12}{\mathrm{C}}^2/\left(\mathrm{N}\cdot {\mathrm{m}}^2\right)}\\ =5.6\times {10}^7\mathrm{V}/\mathrm{m}\end{array}$

Thus, magnitude of $\overrightarrow{E}$in the wall between two layers of charge is $5.6x{10}^{7}V/m$.

The potential energy expression is given as,

$\begin{array}{r}{V}_0=E_{0}d\\ =\left(5.6\times {10}^7\mathrm{V}/\mathrm{m}\right)\left(5.0\times {10}^{-9}\mathrm{m}\right)\\ =0.28\mathrm{V}\end{array}$

The energy density expression is,

$u=\frac{1}{2}{\epsilon }_0{E}_0^2$ …(i)

Volume of the cell is equal to volume of a sphere (assumption),

$V_{\text{cell}}=\frac{4}{3}\pi r^3$

Solve for r,

$\begin{array}{r}r={\left(\frac{3V_{\text{cell}}}{4\pi }\right)}^{\frac{1}{3}}\\ ={\left(\frac{3\left({10}^{-10}{m}^3\right)}{4\pi }\right)}^{\frac{1}{3}}\\ =2.9\times {10}^{-6}\mathrm{m}\end{array}$

The volume of the cell wall with thickness t is given as,

$\begin{array}{r}{V}_{\text{waII}}=4\pi R^{2}t\\ =4\pi {\left(2.9\times {10}^{-6}\mathrm{m}\right)}^2\left(5.0\times {10}^{-9}\mathrm{m}\right)\\ =5.3\times {10}^{-19}{\mathrm{m}}^3\end{array}$

Substitute all the values in equation (i),

$\begin{array}{r}u=\frac{1}{2}\left[8.854\times {10}^{-12}{C}^2/\left(N\cdot m^2\right)\right]{\left[5.6\times {10}^7\mathrm{V}/\mathrm{m}\right]}^2\\ =1.39\times {10}^4\mathrm{J}/{\mathrm{m}}^3\end{array}$

Thus, the total field energy on the cell wall is,

$\begin{array}{r}U=u\times V_{\text{wal}}\\ =\left(1.39\times {10}^4\mathrm{J}/{\mathrm{m}}^3\right)\left(5.3\times {10}^{-19}{\mathrm{m}}^3\right)\\ =7\times {10}^{-15}\mathrm{J}\end{array}$

When air is replaced by dielectric the change in electric field is,

$\begin{array}{r}E=\frac{E_0}{K}\\ =\frac{5.6\times {10}^7\mathrm{V}/\mathrm{m}}{5.4}\\ =1.0\times {10}^7\mathrm{V}/\mathrm{m}\end{array}$

Similarly, for potential difference,

$\begin{array}{r}V=\frac{V_0}{K}\\ =\frac{0.28\mathrm{V}}{5.4}\\ =0.052\mathrm{V}\end{array}$

Thus, the changed electric field and potential difference are $1.0x{10}^{7}V/m$and $0.052V$respectively.