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

Question

3 months ago

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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

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

PART D. What is the direction of the electric field between the walls?

There is no electric field.

Toward the inner wall.

Parallel to the walls.

Toward the outer wall.

PART E. If released from the inner wall, what would be the kinetic energy of a 3fC charge at the outer wall? 1fC=10?15C.

3×10?2J

3×10?8J

3×10?17J

3×10?14J

Physics

1 answer:

serg [3.5K] · Answer 1 · 2026-03-20T02:10:26+03:00

Response:

The provided options are estimations of the precise answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Towards the inner wall

E) $3\times10^{-17}J$

Clarification:

A) The electric field within a parallel plate capacitor is described by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our scenario $\sigma=10^5C/m^2$ and for air, $k=1.00059$ , thus resulting in:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion features one excess elemental charge, hence its charge is $q=1.6\times10^{-19}C$ , and the force that a charge experiences in an electric field E is represented by F=qE, thus we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference across two points that are separated by a distance d in a uniform electric potential E is expressed as $\Delta V=dE$ , leading to:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electric field moves from positive to negative charges, directing it toward the inner wall.

E) The work accomplished by an electric field across a potential difference $\Delta V$ affecting a charge Q is represented by $W=Q\Delta V$ , corresponding to the kinetic energy it receives, yielding:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$