Buggé/Sierzega: Circuits 4
THE PARALLEL PLATE CAPACITOR with a DIELECTRIC (INSULATOR)
In 1837, Michael Faraday found, by experiment, that when 2 capacitors are charged to the same
potential difference, the one that has a dielectric between the plates will have more charge.
Since Q = CV, then the capacitor with the dielectric has a larger capacitance.
Case I)
Battery charges a capacitor, is removed, and then a dielectric is inserted into
the capacitor. (This is aka an isolated capacitor.)
In this case, the charge Q0 on the plates of the capacitor stays the same both before and after
the insertion of the dielectric material.
Q0 +
+
+
+
E0
-
E0
Q0
-
+
+
+
+
-
+
Q’
+
-
E’
E’ is the field within the
dielectric.
Because of the presence
of the dielectric material,
the net electric field (Ed)
decreases.
Quick rule of thumb: When the battery is removed, what happens when
the dielectric is inserted, in order, is that:
E0 decreases (see above) by a factor of κ
∴ V0 decreases by κ
( b/c V = Ed)
Q0 doesn't change! ( b/c there’s no way charge can be added or removed)
∴ C0 increases by κ
( b/c C = Q/V)
We can summarize these results as follows:
(1)
Ed = E0/k
The net field E is weaker with the dielectric in place!
(2)
Vd = V0/k
Whatever happens to E will happen to V b/c V = Ed
(3)
(4)
Since the capacitor is isolated, the charge Q on the capacitor doesn’t change.
Cd = kC0
Cd = (κε0A)/d
k is the dielectric constant, which is based on the material.
(where κ = 1 for air)
Buggé/Sierzega: Circuits 4
Example 1- A parallel plate capacitor has plates of area 100 cm2 that are separated by a
distance 1.0 cm. The capacitor is charged by a 100 V battery, and the battery is then
removed. A dielectric material of dielectric constant, κ = 7.0 is inserted between the
plates. Find:
(a)
the initial capacitance C0
(b)
the "free charge" Q0
(c)
the initial electric field between the plates E0
(d)
the potential difference between the plates with the dielectric in place (Vd)
(e)
the capacitance with the dielectric in place, Cd
Don’t forget the basics!!! You can still use:
C = ε0 A/d
Q = CV
V = Ed
Buggé/Sierzega: Circuits 4
Case II)
Battery charges the capacitor, remains connected, and then a
dielectric material is inserted into the capacitor.
Quick rule of thumb: When the battery is stays in place, what happens when
the dielectric is inserted, in order, is that
V0 doesn't change, because the battery maintains the potential difference.
∴ E0 doesn’t change b/c V0 doesn’t change (V = Ed)
∴ Q0 increases by κ (it has to in order for the battery to maintain the net field E)
∴ C0 increases by κ
( b/c C = Q/V)
E0
Q0 +
+
+
+
++E’
Qd +
+
+
+
+
+
-
V0
V0
Results: As with Case I,
(a)
(b)
Cd = κC0
Since C = Q/V and V is fixed, then
Qd = κQ0
(this is what Faraday found experimentally)
Ed
The field Ed between the plates
is stronger b/c Qd > Q0
Aha!
=
Ed
+
E0
So the net field with the
dielectric in place still
equals the field E0.
E’
Buggé/Sierzega: Circuits 4
Example 9 - A parallel plate capacitor has a capacitance of 22 pF. The plates each have
an area of 50 cm2. A 30 V battery is connected to the capacitor. Find:
(a)
the "free charge" on the plates, Q0
(b)
the electric field, E0
(Answers : 6.6 x 10-10 C, 15,000 N/C )
With the battery still in place, a dielectric of constant κ = 3.5 is now inserted
between the plates. Find:
(c)
Cd
(d)
Qd
(Answers: 77 pF, 2.31 nC )
Buggé/Sierzega: Circuits 4
Access the PhET “Capacitor Lab” Simulation at
http://phet.colorado.edu/en/simulation/capacitor-lab.
1. Go to the “Dielectric” tab. Add a capacitance meter, plate charge meter, and voltmeter to
the simulation. Use the green “Offset” slider to move the dielectric material in between
the plates, and increase the battery voltage to +1.5V.
a. List two quantities that increase with the addition of the dielectric
b. What quantity remains the same?
2. Use the slider to control the dielectric constant in order to establish the following
relationships for a constant voltage:
a. Dielectric constant and capacitance
b. Dielectric constant and charge stored
3. Given the following dielectric constants, which would be (a) the best and (b) the worst
suited to storing the largest possible amount of charge?
!! !
The corresponding equation for capacitance is 𝐶 = !! , where C is capacitance, κ is the
dielectric constant, ε0 is a constant with a value of 8.56 x 10-12 F/m, A is area of the plates, and d
is the distance separating them.
4. Use this formula to find the capacitance for a capacitor with a separation of 1mm of
porcelain between the 10 cm x 10 cm square plates. (Ans: 5.13x10-10 F)
-11
5. A 9x10
F capacitor is made up of two circular plates (r=5cm) separated by 1.0mm.
Find the necessary dielectric constant for this capacitor. (Ans: 1.34)