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

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

Answer 1

Answer:

The options are approximations of the exact answers:

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

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

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

D) Toward the inner wall

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

Explanation:

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

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

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$, and the force a charge experiments under an electric field E is given by F=qE, so we have:

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

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$, so we have:

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

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$, and is equal to the kinetic energy imparted on it, so we have:

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