Question

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.
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

PART B. What is the magnitude of the force on a K+ ion between the cell walls?

2×10?13N

9×10?13N

2×10?11N

3×10?12N

PART C. What is the potential difference between the cell walls?

6×10?3V

1×107V

10V

1×10?2V

Answer 1

Answer:

A) 1×10^6N/C

B) 2×10^-13N

C) 1×10^-2v

Explanation:

For parallel plates,the electric field E is given by:

E=s/eo

Where s= surface charge density

E = 10^-5/8.85×10^-12

E= 1.13×10^6 approximately 1×10^6NC^-2

B) Na has a charge of 1.6×10^-19

F= q×E= (1.13×10^6) × (1.6×10^-19)

F= 1.8×10^-13 approximately 2×10^-13N

C) Potential difference ,V= E×d

d=10nm=10^-8

V= 1.13×10^6 ×10^-8

V = 1.13×10^-2 approximately 1x10^-2v