C
R
Initially the capacitor is uncharged. At time t = 0
the switch is closed and the battery begins to pump
charge from the right capacitor plate to the other
plate.
switch
E
Objective
We want to find the charge on the capacitor
at any time “t” after the switch was closed.
H
H
R
C
+q
-q
i
closed
switch
H
The switch is closed and charge moves through
the battery in the direction shown. At a later
time t the current is i and the charges on the
capacitor plates are +q and -q. The high
potential sides of the capacitor, battery and
resistor are marked.
E
We use Kirchoff’s Rule #2 starting at the green dot and going
clockwise around the loop:
q
∆VR
+E - iR - = 0
C
The current i is also dq . Replacing i we have:
∆VC
dt
q
E - R dq =0
dt C
1
H
H
R
E−R
C
+q
-q
dq E C − q
=
dt
RC
i
closed
switch
H
dq q
− =0
dt C
dq
dt
=
E C − q RC
E
We integrate from an initial charge of zero to the final charge q and
over time from zero to t:
q
t
dq
1
=
∫ E C − q RC ∫ dt
0
0
− [ln ( E C − q )]0q =
H
+q
-q
i
closed
switch
− [ln ( E C − q )]0q =
H C
R
H
E
1
[t ]t0
RC
1
[t ]t0
RC
We can find the charge q(t) at any time t
by inserting the limits and solving:
 EC −q
t
 = −
ln
E
C 
RC

t

−
q (t ) = E C 1 − e RC






As the time becomes large the exponential term decreases to zero so the
final charge on the capacitor is: Q = EC. The charge at any time is also:
t

−
q (t ) = Q 1 − e RC






2
t

−

RC
q (t ) = Q 1 − e






The quantity RC determines how rapidly the
charge on the capacitor increases. It is also
called the time constant: τ = RC (seconds).
The dependence of charge on the time for a 2.0 x10-6 (F) capacitor connected to a
10 (V) battery is shown for different resistances
Charge on Capacitor vs Time
Charge (C)
2.0E-5
final charge Q
τ (s)
0.2
2.0
5.0
R(Ω)
1.0x105
1.0x106
2.5x106
1.0E-5
0.0E+0
0
4
8
12
Time (s)
The current at the time t is the time derivative of the charge:
i (t ) =
t
Q − RC
e
RC
dq
=
dt
=
EC
RC
e
−
t
RC
= Ie
−
t
RC
The current at the initial time t = 0 is I = E/R.
Current in Circuit vs Time
Current (A)
8E-6
For R = 1.0x106 Ω and τ = 2.0 (s).
4E-6
i (t ) = Ie
−
t
RC
0E+0
0
2
4
6
8
10
Time (s)
3
C
R
+Q
-Q
Initially the capacitor has the charges +Q and -Q on
the plates and the switch is open. At time t = 0 the
switch is closed and the positive charge begins to
migrate through the resistor to the other plate
neutralizing some of the negative charge on that
plate.
switch
Objective
We want to find the charge on the capacitor at any
time “t” after the switch was closed.
The Discharging of a Capacitor
H
R
L
H
i
switch
+q
C L
-q
The switch is closed and charge moves in
the direction shown. At a later time t the
current is i and the charges on the
capacitor plates are +q and -q. The high
potential sides of the capacitor and
resistor are marked.
The charge q on the capacitor is decreasing with time so its time
derivative is negative. Therefore the current is the negative
dq
derivative of q since the current must be positive:
i=dt
Kirchoff’s Rule #2 applied around the loop (CW) gives:
+
q
 dq 
− R −
=0
C
 dt 
4
H
R
L
H
i
+
C L
+q
q
 dq 
− R −
=0
C
 dt 
dq
dt
=−
q
RC
-q
q
t
dq
1
∫ q = − RC ∫ dt ⇒
Q
0
switch
q (t ) = Qe
−
q
[ln q]
=−
Q
t
RC
t
RC
The current at any time is:
t
t
i (t ) = I 0 e
−
dq
Q − RC
i (t ) = −
=
e
= I 0 e RC
dt RC
−
t
RC
The graph shows the discharge of a 2.0x10-6 (F) capacitor through a
1.0x106 (Ω) resistor. The initial charge is 5.0x10-6 (C) and the time
constant is: τ = RC = 2.0 seconds.
q (t ) = Qe
Charge vs Time
−
t
RC
Charge (C)
4E-6
Significance of the time constant, τ.
2E-6
When t = τ the value of q is reduced to
the value Q/e or .368Q. For each
successive time constant the value is
reduced by an an additional factor of
1/e.
1.84x10-6 (C)
.677x10-6 (C)
0E+0
0
2
τ
4
6
8
10
Time (s)
5