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snow_lady [41]
4 years ago
9

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

e 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.
PART A. What is the magnitude of the electric field between the membranes?

1×106N/C

1×10?15N/C

5×10?5N/C

9×10?2N/C
Physics
1 answer:
Nina [5.8K]4 years ago
4 0

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

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F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

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So, the ratio of the electrical force to the gravitational force is

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