+
_
+
_
+
_
+
_
uncharged
capacitor
+
_
+
_
+
_
+
_
Initially the switch is open and the capacitor
plates have zero net charge (equal number
of positive and negative charges)
open
switch
battery
The switch is now closed.
charging
capacitor
+
_
+
_
+
_
+
_
+
_
+
_
+
_
+
_
+
+
+
Fc
high
Inside the battery a force Fc ( due to
chemical reactions in the battery) is exerted
on a charge + to move it from the low
potential to high potential side of the
battery.
After this charge reaches the left hand
capacitor plate we have a net positive
charge on that plate and a net negative
charge on the right plate.
closed
switch
low
1
∆Vc
+
+q
∆Vb
high
_
+
-q
q = C∆Vc = C∆Vb
C1
∆V1
Q2
The magnitude of the net final charge on
either plate is:
low
Q1
+
_
The battery will continue to “pump” charge
from the right to the left plate until the
potential difference ∆Vc across the capacitor
equals the potential difference ∆Vb across
the battery.
C2
_
∆V2
The two capacitors C1 and C2 are connected to
the battery in parallel. When the capacitors are
fully charged the potential difference across each
capacitor is the same as the battery potential
difference ∆Vb.
The charges on these capacitors are:
∆Vb
Q1 = C1 ∆V1 = C1 ∆Vb
Q 2 = C 2 ∆V2 = C 2 ∆Vb
The total charge, Qtotal, on both capacitors is also the total charge
which must have passed through the battery during the charging:
Qtotal = Q1 + Q2 = (C1 + C 2 )∆Vb
2
Q1
+
C1
_
∆V1
Q2
C2
It is convenient to introduce an equivalent
capacitor that stores the same total charge when
connected to the same battery. The charge on this
capacitor is:
Q total = C eq ∆Vb
_
+
∆V2
∆Vb
Qtotal
_
+
∆Vb
C1
C eq ∆Vb = (C1 + C 2 )∆Vb
and
Q
∆V1
+
_
∆V2
+
_
C2
Using the total charge from the previous
slide we obtain:
Ceq
Q
∆Vb
C eq = C1 + C 2 + ...
The two capacitors C1 and C2 are connected to
the battery in series. During the charging of
the capacitors a positive charge Q is pumped
by the battery from the bottom plate of C2
(leaving a negative charge of -Q on that
plate). This charge will arrive at the top plate
of C1 (leaving a positive charge of +Q on that
plate).
The charges on the inner plates which are connected together will
redistribute as shown due to the attraction of unlike charges and the
repulsion of like charges. The final charge on each capacitor is the
same even though the capacitors are different.
3
The potential differences across the capacitors are:
C1
Q
∆V1 =
+
_
∆V1
∆V2
+
_
C2
∆Vb
Q
C1
and
∆V 2 =
Q
C2
Q
The sum of the potential differences across both capacitors must
equal the potential difference across the battery:
 1
1
∆Vb = ∆V1 + ∆V2 = Q 
+
 C1 C 2
C1
Q
∆V1
+
_
∆V2
+
_
C2
Q



It is convenient to introduce an equivalent series
capacitor that stores the same charge Q (with same
battery) that each capacitor stores. Note that this is
the charge that was pumped through the battery.
∆Vb The potential difference across the equivalent
capacitor is the same as the potential difference
across the battery:
∆Vb = ∆Veq =
Ceq
∆Veq
Q
C eq
Equating our two values obtained for ∆Vb we have:
Q
+
_
∆Vb
 1
Q
1 

= Q 
+
C eq
 C1 C 2 
1
1
1
=
+
+ ...
C eq C1 C 2
4
Summary for Capacitors in Parallel and Series
Parallel
charges
different
Series
same
∆Vc
same
different
Qeq
Qeq = Qtotal
Qeq = Q1=Q2
Ceq
Ceq =C1 + C2
1/Ceq= 1/C1 + 1/C2
It is not always easy to correctly identify series and parallel
capacitors….
Correct Identification of Series and Parallel Capacitors
The adjacent red and blue capacitors are not in series
The two blue capacitors are in parallel
5
Correct Identification of Series and Parallel Capacitors
These capacitors are in parallel
Correct Identification of Series and Parallel Capacitors
These capacitors are in series
6
Correct Identification of Series and Parallel Capacitors
None of these capacitors are in series or parallel!
7