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

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^(−5) C/m^2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.
(A) What is the magnitude of the electric field between the membranes?

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

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

a) 2×10^13 N
b) 9×10^13 N
c) 2×10^11 N
d) 3×10^12 N

(C) What is the potential difference between the cell walls?

a) 6×10^3 V
b) 1×10^7 V
c) 10 V
d) 1×10^2 V

(D) What is the direction of the electric field between the walls?

a) There is no electric field.
b) Toward the inner wall.
c) Parallel to the walls.
d) Toward the outer wall.

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Dâu 6 months 2021-07-15T20:25:13+00:00 1 Answers 7 views 0

Answers ( )

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    2021-07-15T20:26:29+00:00

    Answer:

    1) A

    2) A

    3) D

    Explanation:

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

    E = σ / ε(o), where

    E = Electric Field

    σ = surface charge density

    E = 10^-5 / 8.85*10^-12

    E= 1.13*10^6 which is approximately 1*10^6 N/C, option A

    B) K has a charge of 1.6*10^-19

    F= q*E= (1.13*10^6) * (1.6*10^-19)

    F= 1.8*10^-13 Which is approximately 2*10^-13 N, option A

    C) Potential difference ,V = Ed

    d = 10 nm= 1*10^-9

    V = 1.13*10^6 * 1**10^-9

    V = 0.0113 v

    V = 1.13×10^-2 which is approximately 1×10^-2v, option D

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