The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of `7.1 xx 10^(-6) C//m^(2)`. What is the magnitude of the electric field within the cell membrane ?

Options

A. `5.9 xx 10^(7) N//C`
B. `8.0 xx 10^(5) N//C`
C. `2.0 xx 10^(3) N//C`
D. `7.6 xx 10^(-1) N//C`

Answer

To find the magnitude of the electric field within the cell membrane, we can use the formula for the electric field \( E \) between the plates of a parallel plate capacitor, which is given by:

\[ E = \frac{\sigma}{\epsilon_0} \]

where:
- \( E \) is the electric field,
- \( \sigma \) is the surface charge density,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step-by-Step Solution:

1. **Identify the surface charge density**:
The problem states that the effective charge density \( \sigma \) on the inner and outer walls of the membrane is:
\[ \sigma = 7.1 \times 10^{-6} \, \text{C/m}^2 \]

2. **Use the formula for electric field**:
Substitute the value of \( \sigma \) into the formula for the electric field:
\[ E = \frac{\sigma}{\epsilon_0} = \frac{7.1 \times 10^{-6}}{8.854 \times 10^{-12}} \]

3. **Calculate the electric field**:
Perform the division:
\[ E = \frac{7.1}{8.854} \times 10^{6 - (-12)} = \frac{7.1}{8.854} \times 10^{6 + 12} = \frac{7.1}{8.854} \times 10^{18} \]
Now, calculating the numerical value:
\[ E \approx 0.801 \times 10^{18} \, \text{N/C} \]
Converting this to scientific notation:
\[ E \approx 8.01 \times 10^{5} \, \text{N/C} \]

4. **Final answer**:
Therefore, the magnitude of the electric field within the cell membrane is:
\[ E \approx 8.01 \times 10^{5} \, \text{N/C} \]