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The semipermeable membrane of a cell is shown, with different concentrations of potassium cations, sodium cations, and chloride anions inside and outside the cell. The ions are represented by small, colored circles. In its resting state, the cell membrane is permeable to potassium and chloride ions, but it is impermeable to sodium ions. By diffusion, potassium cations travel out of the cell, going through the cell membrane and forming a layer of positive charge on the outer surface of the membrane. By diffusion, chloride anions go into the cell, going through the cell membrane and forming a layer of negative charge on the inner surface of the membrane. As a result, a voltage is set up across the cell membrane. The Coulomb force prevents all the ions from crossing the membrane.
The semipermeable membrane of a cell has different concentrations of ions inside and out. Diffusion moves the K + size 12{"K" rSup { size 8{+{}} } } {} and Cl - size 12{"Cl" rSup { size 8{ +- {}} } } {} ions in the direction shown, until the Coulomb force halts further transfer. This results in a layer of positive charge on the outside, a layer of negative charge on the inside, and thus a voltage across the cell membrane. The membrane is normally impermeable to Na + size 12{"Na" rSup { size 8{+{}} } } {} .
This is a graphical representation of a pulse of voltage, or action potential, inside a nerve cell. The voltage in millivolts is plotted along the vertical axis and the time in milliseconds is plotted along the horizontal axis. Initially, between zero and about two point eight milliseconds, the voltage is a constant at about minus ninety millivolts, corresponding to the resting state. Above this section of the graph, a window shows a small cross-section of the cell membrane, with a positively charged outer surface, a negatively charged inner surface, and no ions moving across the membrane. Between two point eight and four point two milliseconds, the voltage increases to a peak of fifty millivolts, corresponding to depolarization of the membrane. A window above this section shows sodium cations crossing the membrane, from outside to inside the cell, so that the membrane’s inner surface acquires a positive charge and its outer surface has a negative charge. Between about four point two and about five point five milliseconds, the voltage drops to a low of about minus one hundred and ten millivolts, corresponding to repolarization of the membrane. A window above this section shows potassium cations crossing the membrane, from inside to outside the cell, so that the membrane’s outer surface again acquires a positive charge and its inner surface has a negative charge. After that, the voltage rises slightly, going back to a constant of about minus ninety millivolts, corresponding to the resting state. This movement of sodium and potassium ions across the membrane is called active transport, and long-term active transport is shown in a window above the final part of the curve.
An action potential is the pulse of voltage inside a nerve cell graphed here. It is caused by movements of ions across the cell membrane as shown. Depolarization occurs when a stimulus makes the membrane permeable to Na + size 12{"Na" rSup { size 8{+{}} } } {} ions. Repolarization follows as the membrane again becomes impermeable to Na + , size 12{"Na" rSup { size 8{+{}} } } {} and K + size 12{"K" rSup { size 8{+{}} } } {} moves from high to low concentration. In the long term, active transport slowly maintains the concentration differences, but the cell may fire hundreds of times in rapid succession without seriously depleting them.

The separation of charge creates a potential difference of 70 to 90 mV across the cell membrane. While this is a small voltage, the resulting electric field ( E = V / d size 12{E = V/d} {} ) across the only 8-nm-thick membrane is immense (on the order of 11 MV/m!) and has fundamental effects on its structure and permeability. Now, if the exterior of a neuron is taken to be at 0 V, then the interior has a resting potential of about –90 mV. Such voltages are created across the membranes of almost all types of animal cells but are largest in nerve and muscle cells. In fact, fully 25% of the energy used by cells goes toward creating and maintaining these potentials.

Electric currents along the cell membrane are created by any stimulus that changes the membrane’s permeability. The membrane thus temporarily becomes permeable to Na + size 12{"Na" rSup { size 8{+{}} } } {} , which then rushes in, driven both by diffusion and the Coulomb force. This inrush of Na + size 12{"Na" rSup { size 8{+{}} } } {} first neutralizes the inside membrane, or depolarizes it, and then makes it slightly positive. The depolarization causes the membrane to again become impermeable to Na + size 12{"Na" rSup { size 8{+{}} } } {} , and the movement of K + size 12{"K" rSup { size 8{+{}} } } {} quickly returns the cell to its resting potential, or repolarizes it. This sequence of events results in a voltage pulse, called the action potential . (See [link] .) Only small fractions of the ions move, so that the cell can fire many hundreds of times without depleting the excess concentrations of Na + size 12{"Na" rSup { size 8{+{}} } } {} and K + size 12{"K" rSup { size 8{+{}} } } {} . Eventually, the cell must replenish these ions to maintain the concentration differences that create bioelectricity. This sodium-potassium pump is an example of active transport , wherein cell energy is used to move ions across membranes against diffusion gradients and the Coulomb force.

The action potential is a voltage pulse at one location on a cell membrane. How does it get transmitted along the cell membrane, and in particular down an axon, as a nerve impulse? The answer is that the changing voltage and electric fields affect the permeability of the adjacent cell membrane, so that the same process takes place there. The adjacent membrane depolarizes, affecting the membrane further down, and so on, as illustrated in [link] . Thus the action potential stimulated at one location triggers a nerve impulse that moves slowly (about 1 m/s) along the cell membrane.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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