5C20.10 Parallel Plate Capacitor
Dielectrics
Abstract
A voltage is placed across a parallel plate capacitor by using the electrophorus to put a charge on one side of
the capacitor. The presence of the charge can be seen on the electroscope. Once charged, a dielectric material
can be placed between the plates of the capacitor. The electric field between the two plates causes an induced
electric field of opposite polarity across the dielectric. The net electric field is reduced causing a decrease in the
voltage across the capacitor. This change in the voltage can be detected by the electroscope.
Picture
Setup
Setup is 5 minutes
Safety Concerns
None.
Equipment
• Dielectric apparatus
• Fur
• Transparent electroscope
• Electrophorus
• Circular metal plate
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Procedure
Connect one of the parallel plates to the black ring on the transparent electroscope and the other to the outer
ring of the electroscope. Adjust the gap between the plates so that the dielectric can be easily inserted into the
gap. Remove dielectric. Rub the electrophorus with the fur, place the circular metal plate on the electrophorus
and press down on it with a finger. There should be a slight electrical discharge. Pickup the circular plate and
touch it to the top plate of the capacitor at which time there should be another electrical discharge and the needle
of the electroscope should have deflected slightly from center. Repeat charging the upper plate until the needle
of the electroscope is near its maximum displacement. Once charged return the dielectric into the gap between
the parallel plates at which time the electroscope should show a decrease in the voltage across the capacitor.
On humid days the electrophorus may not work very well. If this happens try using the Wimshurst generator
to provide the required voltage. Attach the sides of the Whimshurst’s leyden jar capacitor different sides of the
parallel plate capacitor using the leads and the alligator clips. Spin the generator crank in the clockwise direction
to charge up the plates. When finished use the discharging rod to discharge the plates.
When complete, place the dielectric between the plates and adjust the vertical positions of the plates so that
they are as close to the dielectric as possible while still allowing the dielectric to be easily removed. Rotate the
plates so that are aligned with each other while still covering the entire surface of the dielectric.
Stack electrophorus and circular disk on base of dielectric apparatus and place in cupboard. Return fur to box
of furs, and return electroscope to the cupboard with the rest of the electroscopes.
Theory
The electric field between the two parallel plates of a capacitor, which have equal and opposite charge densities
of σ, is illustrated in Figure 1a and is described by the equation,
Eo =
σ
,
o
(1)
where Eo is the electric field between two parallel plates with no dielectric, and o is permittivity of free space
(o = 8.85x10 − 12C 2 /N m2 ).
When a dielectric is placed between the
plates of the capacitor the electric field will
+σ
−σ
induce a charge density of magnitude σi ,
Eo
E
–
on the sides of the dielectric. A dielectric is
–
+
+
an insulator and therefore the charges in it
–
+ –
+ –
+
are bound to the atoms or molecules. The
charges are not free to move around the
–
–
+
+
dielectric, they can only be shifted within
each atom or molecule. As a result the
–
–
–
+
+
+
induced charge density is less than charge
density of the plates of the capacitor.
–
–
+
+
Since the polarity of the induced charge
–
+ –
+ –
+
density will be opposite the polarity of the
capacitor as shown in Figure 1b, part of
–
–
+
+
the charge density of the capacitor will be
neutralized. A lower net charge density will
+σi
−σi
+σ
−σ
resulting in a smaller electric field across
the capacitor.
(a)
(b)
The electric field across a parallel plate
Figure 1: Diagram showing the electric field lines between parallel capacitor with a dielectric is described by,
plates without dielectric material in Figure 1b and with dielectric
σ − σi
material in Figure 1a. The charge densities within the plates, σ and
E=
,
(2)
o
the induced charge densities within the dielectric material, σi are
also labelled.
where E is the electric field across the capacitor with the dielectric in place. The
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ratio of the electric field across a capacitor without any dielectric to the electric field across the same capacitor
with a dielectric is a characteristic of the material of which the dielectric was made. This constant is obtained by
taking the ratio of Equations 1, and 2. The resulting expression is
Eo
σ
=
= K,
E
σ − σi
(3)
where K is the dielectric constant. Note that the induced charge density cannot be larger that the original charge
density, or rather σi ≤ σ, therefore K ≥ 1.
The voltage across a parallel plate capacitor is given by the expression,
V = Ed,
(4)
where d is the distance between the two plates of the capacitor. Substituting Equation 4 into Equation 3 reveals
the expression,
Vo
,
(5)
V =
K
where Vo is the initial voltage across the capacitor without the dielectric, and V is the final voltage across the
capacitor with the dielectric in place.
Equation 5 gives the relationship between the initial and final voltages across the capacitor. Since K ≥ 1 the
final voltage must always be less then or equal to the initial voltage. This prediction is confirmed by the reaction
of the electroscope when a dielectric is placed between the plates of a charged capacitor.
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References
[1] Freier, G. D. and Anderson, F. J. A Demonstration Handbook for Physics, ”Ed-2 Dielectrics”, American
Association of Physics Teachers One Physics Ellipse, College Park MD, 1996. pg E-19.
[2] Sutton, Richard Manliffe. Demonstration Experiments in Physics, ”E70. Dielectric Constant and Capacitance”,
McGraw-Hill Book Company Inc., New York and London, 1938. pg 277.
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