Question

A typical cell has a membrane potential of $-70\mathrm{mV}$, meaning that the potential inside the cell is $70\mathrm{mV}$ less than the potential outside due to a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outer surface. This effectively makes the cell wall a charged capacitor. Because a cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor. How much energy is stored in the electric field of a $50\mu m$diameter cell with a $7.0nm$thick cell wall whose dielectric constant is $9.0$?

Short answer

The amount of energy is stored in the electric field of a $50\mu m$diameter cell with a $7.0nm$thick cell wall with dielectric constant $9.0$is$0.22pJ$

Step-by-step solution

Given Information

Potential energy outside$=-70mV$

Potential energy inside$=70mV$

Diameter$=50\mu m$

Thickness$=7.0nm$

Dielectric constant$=9.0$

Explanation

A parallel-plate capacitor is a charge that is applied at a potential difference $∆V$, whose plates are separated by a distance $d$, and whose area is the same as that of a sphere with a diameter $D$. We need the energy this capacitor is storing.

The energy stored in any capacitor of capacitance $C$charged to potential difference $\Delta V$will be given by

$U=\frac{C(\Delta V{)}^2}{2}$

Find the capacitance,

Since we assumed a parallel plate model, then it will be given by

$C=\epsilon {\epsilon }_0\frac{A}{d}$

where $\epsilon$is the dielectric constant of the medium. The surface of a sphere of diameter $A$is given by

$A=4\pi R^2=\pi D^2$

Substituting, we find the capacitance to be

$C=\epsilon {\epsilon }_0\frac{\pi D^2}{d}$

Substituting the expression for the capacitance we found at the expression for the energy, we find

$U=\pi \epsilon {\epsilon }_0\frac{D^2(\Delta V{)}^2}{2d}$

In the given numerical case, we will have

localid="1648647894520" $U=\left(\frac{\pi \times 9\times 8.85·{10}^{-12}{\left(5\times {10}^{-5}\mu m\right)}^2\times 0.{07}^{2}mV}{2\times 7·{10}^{-9}m}\right)$

Multiply the values,

$=2.2·{10}^{-13}\mathrm{J}$

$=0.22\mathrm{pJ}$

Final Answer

Hence, the amount of energy is stored in the electric field is$0.22pJ.$