Experiment 16
Capacitance and the nerve cell
Purpose
To try to understand the ready state of the nerve cell by studying properties of capacitors.
Apparatus
Electrometer, High voltage source, Proof plane, Air capacitor. Wires: Alligator clip to BNC connector
with resistor at alligator clip, Ground wire with resistor between high voltage source and electrometer.
Nerve Cell
From our experiment where we plotted electric field lines and our study of the nerve cell, we
should understand that charges reside on either side of the nerve cell membrane. These charges create an
electric field that starts from the positive side of the membrane and moves to the negative charges. From
the definition of electric field strength [equation 6 from experiment 12] and using the potential difference
and the separation distance between charges, we find that the electric field that passes through the
membrane is approximately equal to 8 x 10 6 V/m. This is a very large field strength. An electric field in air
can cause sparking to occur if the field is greater than 3 x 10 6 V/m.
It is important to learn how this configuration of charges and the electric field produced can send a
stimulus to the brain. The first part of this learning process is to realize the nerve cell is storing charges on
either side of its membrane. To a physicist a capacitor does the same thing.
Capacitance
A capacitor is a charge storage device. A capacitor consists of two metal plates that are parallel to
each other and separated by a distance. A positive charge is placed on one of the plates while an equal
amount of negative charge is drawn up from the earth to charge the other plate. The separation distance
between the plates ensures that no charge can flow from one plate to another. In this way, a capacitor can
store a charge that has been placed on it.
A relationship exists between the amount of charge that can be put on a capacitor and the electrical
potential difference between the plates. Theory and experiments have shown that the charge and potential
difference are directly proportional by a constant.
q = CV
Eq. 1)
Equation one illustrates the relationship between the charge and the electrical potential difference.
In this equation q represent the charge placed on the plates measured in coulombs, V is the electrical
potential difference measured in volts, and C is the proportionality constant called the capacitance. The unit
of capacitance is the farad. Typical measurements of capacitance in electronic devices are from microfarads
(1F = 10–6 F) to a picofarad (1 pF = 10-12 F).
The capacitance of the parallel plates is affected by the geometry of the system. From a detailed
analysis of point charges lying on the surface of the plates, and the rules of vector addition, it can be shown
that the electric field between the plates is equal to
E=
q
o A
.
Eq. 2)
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Here E is the strength of the electric field, q is the amount of charge on the plates, A is the cross sectional
area of the plates and o is the permittivity of free space (remember o, is equal to 8.85x10-12 C2/Nm2). The
permittivity of free space is a constant. We also know from an earlier experiment that the electric field
between the plates will be equal to
E=−
V
d
.
Eq. 3)
In this case E is the electric field strength, V is the potential difference between the plates, and d is the
distance between the plates. By combining equations 2 and 3 and then substituting equation 1 for the
charge on the plates we arrive at
C=
o A
d
.
Eq. 4)
This shows us that the capacitance value relates the amount of charge and the potential difference.
We know from previous experiments what the electric field looks like between two parallel lines
of opposite charges. This is a two dimensional view of how the electric field looks between the plates of a
parallel plate capacitor. We can increase the capacitance of a capacitor by placing a substance called a
dielectric between the plates.
Dielectrics are insulating materials. The material prevents charges from flowing across them from
one plate to the other. The overall effect of the dielectric when placed between two charged plates is to
reduce the electric field between the plates. If we take a close look at the dielectric within an electric field
we see that the dielectric becomes polarized. This means that the charge distribution about the atoms and
molecules that makes up the dielectric begin to separate. The electrons that surround the atoms do not leave
it but are pulled to one side of the atom thereby giving that side of the atom a more negative charge than the
opposite side. Some of the electric field lines will then move from the charge plates to the negative side of
the atom and some of the electric field lines will pass all of the way through the material. Different
dielectric materials will affect the electric field by different magnitudes. We use the ratio of electric field
strengths to define the dielectric constant of a material as
k=
Eo
E
Eq. 5)
Here k is the dielectric constant, Eo is the electric field strength of an empty capacitor and E is the strength
of the electric field inside the dielectric. It should be obvious that k will be equal to or greater than 1. We
can use this ratio to define the magnitude of the electric field inside the capacitor in equation 2 and by
combining with equation 3 rewrite equation 4 as
C=
ko A
d
Eq. 6)
The dielectric constant increases the capacitance of equation 4 by a factor of k.
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Figure 1.) The illustration on the left is the electric field inside a capacitor with a dielectric. The molecules in the
dielectric become polarized and decrease the electric field passing through the center of the dielectric. The figure on
the right illustrates the electric field of an empty capacitor.
Nerve cell and the capacitor
From the discussion above, it should be obvious that the nerve cell when in its ready state is just a
charged capacitor. Also, from the information above, we should be able to calculate the capacitance and the
dielectric constant of the neuron and its membrane. Careful studies of the nerve cell ion concentrations
show us that approximately 3.5 x 10 -13C of charge exist on the membrane. Using this and the potential
difference across the membrane gives the cell a capacitance of approximately 5x10 -12 F or 5 pF. The typical
nerve cell has a volume of 1 x 10-15 m3 and a surface area of about 5 x 10-10 m2. We can use this in equation
6 and find the dielectric constant to be equal to 10.2. It is because of the high dielectric constant that the
field strength can be so high. Later, we learn more about how capacitors and the nerve cell handle the
charging and discharging of themselves.
The experiment
Hints and suggestions:
The students should familiarize themselves with the electrometer scale. Use the zero to 10 scale.
The students should make all measurements from the same location relative to the capacitor plates.
The position of the students’ bodies change the capacitance of the system so arms and torsos should return
to the same far away location at the moment each measurement is read.
Part 1
Capacitance varied, charge held constant, voltage measured
In this part of the experiment we will see how the capacitance is affected by the geometry of the
plates. We will place a constant quantity of charge on the plates of the capacitor and measure the
capacitance as we change the distance between the plates. Figure 2 illustrates the equipment setup for this
part of the experiment. You are also required to derive an equation before coming to class as described on
the next page.
Connect the voltage leads of the electrometer across the capacitor. We need to ensure that no
excess charge remains on the capacitor. We can ground the capacitor and remove all excess charge by
depressing the “zero check” switch on the electrometer. Connect the green binding post of the power
supply to the sphere and set the power supply to 1000 VDC. Set the capacitor plate separation to 3 mm.
Important: Please keep the sphere and power supply and capacitors at opposite ends of the table. If they
are too close static charge can leak off the sphere onto the capacitor. Please also keep the sphere and power
supply away from the computer. Large amounts of static electricity can kill a computer chip.
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Now touch the proof plane to the high voltage sphere and then touch it to the ungrounded plate of
the capacitor. You should see noticeable movement in the voltage read from the electrometer gauge. The
electrometer will measure the voltage across the capacitor. It may be necessary to ground the plates of the
capacitor to ensure that only one unit of charge has been placed on the capacitor. We will separate the
plates of the capacitor slowly and record the voltage across the plates and the separation distance of the
plates. Start at 3 mm of separation and take the voltage measurements every mm up to 11 mm. Do not
charge the capacitor more than once in this part of the experiment.
In preparation for this lab derive the expected slope and intercept of a plot of 1/V vs. 1/d using
equations 1) and 4) with an additive constant representing the extraneous capacitance. A graph will be
made with your data of 1/voltage vs. 1/distance between the plates. You should be able to calculate the
charge placed on the plates and the extraneous capacitance from this graph using the quantities of slope,
intercept and the area of the plates. You should derive the equations for the charge and extraneous
capacitance in terms of the slope and intercept before coming to class so that the in-class calculation can be
quickly accomplished.
Hint: It is important to be motionless accept for the change in the plate separation and to make the
entire run in under one minute. One person calls out the millimeter separation while moving the plates
while the other person writes down the voltages.
Sphere
Proof Plane
Capacitor
High voltage source
Electrometer
.
Figure 2. This illustrates the equipment setup. The high voltage source is connected to a conducting sphere. A proof
plane is used to transfer units of charge from the conducting sphere to the capacitor. The electrometer will measure the
voltage across the capacitor.
Part 2
Capacitance held constant, charge varied, voltage recorded
In this part of the experiment we will explore how the voltage and charge are related to each other
on a capacitor. We will do this by measuring the voltage across an open-air capacitor as uniform amounts
of charge are placed upon the capacitor.
With an initial plate separation of 3 mm., use the proof plane to transfer charge from the sphere to
the ungrounded capacitor plate (i.e., the plate connected to the red electrometer lead). The charge is
transferred by touching the proof plane to the sphere and then to the capacitor in the same manner each
time. This will ensure equal amounts of charge are placed on the capacitor each time the capacitor is
touched by the proof plane.
Please note that we know from part 1 the approximate amount of charge deposited by each touch
of the proof plane, measured in coulombs, which will be deposited on the capacitor at this time. By
touching the proof plane to the sphere and then the capacitor, we have placed 1X amount of charge on the
capacitor. After we have touched the sphere and capacitor a second time, we will have 2X amount of
charge on the capacitor. After the third time we have touched the sphere and capacitor, the capacitor has 3X
amount of charge on it, etc. In each of these cases X represents the constant amount of charge that is being
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placed upon the capacitor. The value of X should be equal to the value of charge placed upon the capacitor
in part 1 if all is done precisely the same as in part 1. Use the value of charge in part 1 to calculate the
charge placed upon the capacitor.
Record the voltage of the capacitor and the amount of charge placed upon the capacitor. Construct
a graph of the amount of charge placed on the capacitor vs. the voltage of the capacitor (q vs. V). The graph
should be a straight line and the slope of this line should be equal to the value of the total capacitance of the
capacitor and surroundings. Calculate the difference between the expected parallel plate capacitance from
equation 4) and the slope you found from your data. This is the extraneous capacitance of the system due to
the wires and the electrometer. This value you calculated for the extra capacitance is used in part 3. Also,
compare this extraneous capacitance to that calculated in part 1.
Part 3
Dielectric and the change in capacitance.
In this part of the experiment we would like to see the effect of a dielectric placed in-between the
plates of the capacitor. Using the same set-up as in part 2, with the plates 3 mm. apart, insert the dielectric
material between the plates of the capacitor. Record the voltage and the amount of charge that you have
placed upon the capacitor. Construct a new graph as you did in part two. This time note the slope of the
graph. Is the slope of the graph greater than or less than the slope in part two? Remember that the slope of
the line is equal to the capacitance of the capacitor.
We would like to determine the dielectric coefficient. In order to do this we must use the
capacitance of the electrometer and the wires connected to it. The value of the extra capacitance was found
at the end of part 2. From the data collected in part two, determine the average voltage change between unit
charge depositions. Do the same for the graph just made in part 3. With the distance between plates set to 3
mm, determine the capacitance for this capacitor with no dielectric in it. You will need to use equation 4
just as you did in part 1. The dielectric coefficient can be calculated by using the following equation
k=
C1 (Vo − V1 ) + C2Vo
C2V1
Eq. 7)
In this equation k is the dielectric coefficient, C1 is the capacitance of the electrometer and the wires (the
value calculated in parts 1) & 2), C2 = oA/d is the capacitance of the open air capacitor, V o is the average
voltage change from part 2 and V1 is the average voltage change for part 3. Compare this value with that
from the dielectric listed here: (1.5).
Question
1.
Why, in part one of this experiment, do you only need to touch one of the plates of the capacitor with
the proof plane?
2.
How large of a base charge (X) did you place on the capacitor?
3.
What is the dielectric constant of the material used in this experiment?
4.
Go through the calculations necessary to calculate the capacitance of the neuron membrane. Use the
membrane thickness given in the previous lab while using the charge and potential difference given in
this lab.
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