A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10-5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.

What is the magnitude of the electric field between the membranes?

a. 1×10^-6N/C
b. 1×10^-15N/C
c. 5×10^-5N/C
d. 9×10^-2N/C

The magnitude of the electric field between the membranes is  1.13 x 10⁶ N/C

Explanation:

Given;

the distance of separation of the parallel plate capacitor, d =  10nm

the charge density, σ = 10⁻⁵C/m²

the magnitude of the electric field between the membranes, is calculated using the formula below;

E = σ / ε₀

Where;

ε₀ is permittivity of free space, = 8.85 x 10⁻¹² C²/Nm²

E is magnitude of the electric field between the membranes

σ is surface charge density

E = (10⁻⁵) / (8.85 x 10⁻¹²)

E = 1.13 x 10⁶ N/C

Therefore, the magnitude of the electric field between the membranes is  1.13 x 10⁶ N/C

a, c, d, e are all applying about success strategies.

if they have a finite resultant