Download Slide 1

Today’s agenda:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
Capacitors in Circuits
Recall: this is the symbol representing a
capacitor in an electric circuit.
And this is the symbol for a battery…
…or this…
…or this.
+-
Circuits Containing Capacitors in Parallel
Vab
Capacitors connected in parallel:
C1
a
C2
b
C3
+ V
The potential difference (voltage drop) from a to b must equal V.
Vab = V = voltage drop across each individual capacitor.
Note how I have introduced the idea that when circuit components are connected in parallel, then the voltage
drops across the components are all the same. You may use this fact in homework solutions.
C1
Q=CV
 Q1 = C1 V
& Q2 = C2 V
& Q3 = C3 V
Q1
a
C2 -
+
Q2
C3
Q3
+ V
Now imagine replacing the parallel
combination of capacitors by a single
equivalent capacitor.
a
By “equivalent,” we mean “stores the same
total charge if the voltage is the same.”
Qtotal = Ceq V = Q1 + Q2 + Q3
Important!
Ceq
Q
+ V
Summarizing the equations on the last slide:
Q1 = C1 V
Q2 = C2 V
Q3 = C3 V
C1
C2
a
Q1 + Q2 + Q3 = Ceq V
C3
+ -
Using Q1 = C1V, etc., gives
V
C1V + C2V + C3V = Ceq V
C1 + C2 + C3 = Ceq
Generalizing:
(after dividing both sides by V)
Ceq = Ci
(capacitors in parallel)
b
Circuits Containing Capacitors in Series
Capacitors connected in series:
C1
C2
C3
+ -
+Q V -Q
An amount of charge +Q flows from the battery to the left plate
of C1. (Of course, the charge doesn’t all flow at once).
An amount of charge -Q flows from the battery to the right
plate of C3. Note that +Q and –Q must be the same in
magnitude but of opposite sign.
The charges +Q and –Q attract equal and opposite charges to
the other plates of their respective capacitors:
C1
+Q -Q
A
C2
+Q -Q
B
C3
+Q -Q
+ V
These equal and opposite charges came from the originally
neutral circuit regions A and B.
Because region A must be neutral, there must be a charge +Q
on the left plate of C2.
Because region B must be neutral, there must be a charge -Q
on the right plate of C2.
Vab
a
C1
+Q -Q
V1
C2
A
+Q -Q
V2
C3
B
b
+Q -Q
V3
+ V
The charges on C1, C2, and C3 are the same, and are
Q = C1 V1
Q = C2 V2
Q = C3 V3
But we don’t know V1, V2, and V3 yet.
We do know that Vab = V and also Vab = V1 + V2 + V3.
Note how I have introduced the idea that when circuit components are connected in series, then the voltage
drop across all the components is the sum of the voltage drops across the individual components. This is
actually a consequence of the conservation of energy. You may use this fact in homework solutions.
Let’s replace the three capacitors by a single equivalent
capacitor.
Ceq
+Q -Q
V
+ V
By “equivalent” we mean V is the same as the total voltage
drop across the three capacitors, and the amount of charge Q
that flowed out of the battery is the same as when there were
three capacitors.
Q = Ceq V
Collecting equations:
Q = C1 V1
Q = C2 V2
Q = C3 V3
Important!
Vab = V = V1 + V2 + V3.
Q = Ceq V
Substituting for V1, V2, and V3:
Q
Q
Q
V=
+
+
C1 C 2 C 3
Substituting for V:
Q
Q
Q
Q
=
+
+
Ceq C1 C2 C3
Dividing both sides by Q:
1
1
1
1
=
+
+
Ceq C1 C2 C3
Generalizing:
OSE:
1
1
=
Ceq
Ci
i
(capacitors in series)
Summary (know for exam!):
Series
C1
C2
same Q, V’s add
C3
1
1

Ceq
i Ci
C1
Parallel
C2
C3
same V, Q’s add
Ceq   Ci
i