A cell membrane is a thin layer enveloping a cell. The thickness of the membrane is much less than the size of the cell. In a static situation the membrane has a charge distribution of −2.5×10⁻⁶ C/m² on its inner surface and +2.5×10⁻⁶ C/m² on its outer surface. Draw a diagram of the cell and the surrounding cell membrane. Include on this diagram the charge distribution and the corresponding electric field. Is there any electric field inside the cell? Is there any electric field outside the cell?

Electric flux is a scalar quantity defined as the dot product of electric field vector to the area vector. Mathematically,

$\mathit{\phi }\mathbf{=}\underset{\mathbf{c}}{\mathbf{\oint }}\overrightarrow{\mathbf{E}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{s}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{)}$

Here,$\overrightarrow{\mathbf{E}}$ is the electric field vector and$\mathit{d}\overrightarrow{\mathbf{s}}$ is the area vector.

From Gauss law, the electric flux is,

$\mathit{\phi }\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}}{{\mathbf{\epsilon }}_{\mathbf{\circ }}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{)}$

Here,${\mathit{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$ is the charge enclosed inside the hypothetical Gaussian surface, and${\mathit{\epsilon }}_{\mathbf{\circ }}$ is the permittivity of the free space.

From equation (1.1) and (1.2),

$\underset{\mathbf{c}}{\mathbf{\oint }}\overrightarrow{\mathbf{E}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{s}}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}}{{\mathbf{\epsilon }}_{\mathbf{\circ }}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{)}$

Consider a hypothetical Gaussian surface inside the cell. The charge enclosed inside this hypothetical surface is zero $(q_{enc}=0)$. The electric field inside the cell is given by equation (1.3).

Substituting all known values,

$$\begin{gathered} \underset{c}{\oint }\overrightarrow{E}·d\overrightarrow{s}=\frac{0}{{\epsilon }_{\circ }} \\ =0 \end{gathered}$$

Hence, the electric field inside the cell is zero.

Consider a hypothetical Gaussian surface outside the cell. The charge enclosed inside this hypothetical surface is zero (${\mathit{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}\mathbf{=}\mathbf{0}$, the charge due to positive charge distribution cancels the charge due to negative charge distribution as they have equal magnitude). The electric field outside the cell is given by equation (1.3).

Substituting all known values,

$$\begin{gathered} \underset{c}{\oint }\overrightarrow{E}·d\overrightarrow{s}=\frac{0}{{\epsilon }_{\circ }} \\ =0 \end{gathered}$$

Hence, the electric field outside the cell is zero.

The electric field due to positively charged cell membrane is,

$E^{+}=\frac{{\sigma }^{+}}{2{\epsilon }_{\circ }}$

Here, ${\mathit{\sigma }}^{\mathbf{+}}$is the surface charge density of the cell membrane $\left({\sigma }^{+}=2.5\times {10}^{-6}C/m^2\right)$, and ${\mathit{\epsilon }}_{\mathbf{\circ }}$is the permittivity of the free space $\left({\epsilon }_{\circ }=8.854\times {10}^{-12}F/m\right)$.

The electric field due to negatively charged cell is,

$E^{-}=\frac{{\sigma }^{-}}{2{\epsilon }_{\circ }}$

Here,${\sigma }^{-}$ is the surface charge density of the cell membrane $\left({\sigma }^{-}=-2.5\times {10}^{-6}C/m^2\right)$, and${\epsilon }_{\circ }$ is the permittivity of the free space $\left({\epsilon }_{\circ }=8.854\times {10}^{-12}F/m\right)$.

The net electric field between the cell membrane and the cell is,

$$\begin{gathered} E=E^{+}-E^{-} \\ =\frac{{\sigma }^{+}}{2{\epsilon }_{\circ }}-\frac{{\sigma }^{-}}{2{\epsilon }_{\circ }} \end{gathered}$$

Substituting all known values,

$$\begin{gathered} E=\frac{\left(2.5\times {10}^{-6}C/m^2\right)}{2\times \left(8.854\times {10}^{-12}F/m\right)}-\frac{\left(2.5\times {10}^{-6}C/m^2\right)}{2\times \left(8.854\times {10}^{-12}F/m\right)} \\ =2.82\times {10}^{5}N/C \end{gathered}$$

Hence, the electric field between the cell membrane and cell is $2.82\times {10}^{5}N/C$.