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Vinil7 [7]
3 years ago
15

Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistanc

e (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?

Physics
1 answer:
Korolek [52]3 years ago
6 0

Answer:

<em>check the diagram in the attachment below.</em>

After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the  capacitor will be short circuited.

Explanation:

question

Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?

ANSWER;

After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the  capacitor will be short circuited.

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a) 8.93\cdot 10^{-30} N, downward

b) 1.76\cdot 10^{-17} N, upward

c) 1.20\cdot 10^{-17} N, downward

Explanation:

a)

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m=9.11\cdot 10^{-31} kg is the mass of the electron

g=9.8 m/s^2 is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N

And the direction is downward, towards the Earth's centre.

b)

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

F=qE

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

q=1.6\cdot 10^{-19} C is the charge of the electron

E=110 N/C is the strength of the electric field

Substituting,

F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N

And since the electron has negative charge, the  direction of the force is opposite to that of the electric field, so upward.

c)

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

F=qvB

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

q=1.6\cdot 10^{-19} C is the charge of the electron

v=1.50\cdot 10^6 m/s is the velocity

B=50.0\mu T = 50.0 \cdot 10^{-6} T is the strength of the magnetic field

Substituting,

F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N

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- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

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