Question (2jg0001)
Charging Capacitor in an RC circuit
An RC circuit has a 6-V battery, a
the capacitor plates. At
t = 0,
1 kΩ
resistor, and a
100 µF
capacitor. At
t = 0,
there is no charge on
the switch is closed, and the capacitor begins to charge.
Figure 1: RC Circuit
(a) What is the time constant of the charging capacitor?
(b) Sketch graphs of
(c) At
t = 0.15 s,
Q
vs.
what is
t, ∆VC
vs.
t, ∆VR
Q, ∆VC , ∆VR ,
and
vs.
~
E
and
I
vs.
t.
I?
(d) Sketch the charge on the capacitor plates at
(e) Sketch
t,
t = 0.15 s.
due to surface charge on the wire at point P at
t = 0.15 s.
(f ) Sketch the fringe eld due to surface charge on the capacitor plates, at point P at
(g) Sketch
~ net
E
at point P at
t = 0.15 s.
t = 0.15 s.
Solution
(a)
The time constant for an RC circuit is
τ
(b)
= RC
=
(1 × 103 Ω)(100 × 10−6 F)
=
0.1 s
(1 − e−t/τ ), asymptotically approaching
= (100 × 10−6 F)(6 V) = 6 × 10−4 C.
The capacitor charges, so Q increases in proportion to
maximum possible charge which is
Qmax = C∆VC,max
Q = Qmax (1 − e−t/τ )
Q =
(6 × 10−4 C)(1 − e−t/τ )
the
Figure 2: Q vs. t for a charging capacitor
Since
Q = C∆VC ,
as charge builds up on the plates, the voltage across the capacitor increases proportion-
ately, approaching the maximum possible voltage, 6 V.
∆VC
=
∆VC,max (1 − e−t/τ )
∆VC
=
(6 V)(1 − e−t/τ )
Figure 3: Voltage vs. t for a charging capacitor
The voltage across the resistor is found by applying Conservation of Energy to a loop around the circuit.
∆VC + ∆VR
=
∆Vbat
∆VR
=
∆Vbat − ∆VC
∆VR
=
6 − 6(1 − e−t/τ )
∆VR
=
6 − 6(1 − e−t/τ )
∆VR
=
6(1 − 1 + e−t/τ )
∆VR
=
6e−t/τ
Note that this result simply means that as the capacitor voltage increases to 6 V, the resistor voltage starts
at 6 V and exponentially decays to zero. A graph is shown below.
Figure 4: Voltage vs. t for the resistor in an RC circuit where the capacitor is charging.
The current through the resistor is proportional to the voltage across the resistor, according to Ohm's law
(∆VR
= IR).
Therefore, if the voltage across the resistor exponentially decays, the current also exponentially
decays. The maximum current is found from Ohm's law
∆VR,max
Imax
Imax
Imax
= Imax R
∆VR,max
=
R
6V
=
1000 Ω
= 0.006 A
Thus, the current as a function of time is
I
and the graph is shown below.
=
(0.006 A)e−t/τ
Figure 5: Current vs. t for a charging capacitor.
(c) Qmax = C∆VC,max = (100 × 10−6 F)(6 V) = 6 × 10−4 C. At t=0.15 s,
Q = Qmax (1 − e−t/τ )
0.15 s
= (6 × 10−4 C)(1 − e− 0.1 s )
= 4.66 × 10−4 C
∆VC,max = 6 V.
The voltage across the capacitor
∆VC
at t=0.15 s
0.15 s
=
∆VC,max (1 − e− 0.1 s )
=
4.66 V
Conservation of Energy applied to the circuit gives
∆VR
Ohm's Law is
∆VR = IR.
Thus, the current
I
is
∆VC + ∆VR = ∆Vbat .
=
∆Vbat − ∆VC
=
=
6 − 4.66
1.33 V
at t=0.15 s
=
=
=
is
∆VR
R
1.33
1000
1.33 × 10−3 A
Thus
at t=0.15 s,
(d)
To sketch the charge on the plates at
t = 0.15 s,
consider the direction of the electron current which
ows through the circuit away from the negative terminal of the battery and toward the positive terminal
of the battery as shown below.
Figure 6: Electron current as capacitor is charging.
Electrons will pile up on the lower" plate in the circuit diagram and will leave the upper" plate in the
circuit diagram, as shown below.
Figure 7: Charge building up on the capacitor plates.
(e)
To sketch
~
E
due to the surface charge gradient on the wire at point P at
of conventional current at point A.
conventional current.
~
E
t = 0.15 s, consider the direction
due to the surface charge gradient on the wire is in the direction of
Figure 8: Electric eld due to the surface charge gradient on the wire at point P.
(f )
To sketch the fringe eld due to the charged capacitor plates at point P, note that the net electric eld
due to the plates will point away from the positively charged plate since it is closer to point P than the
negatively charged plate and will thus dominate. The fringe eld will be smaller than the electric eld due
to the surface charge gradient since the net electric eld is in the direction of the conventional current, I;
after all, at
t = 0.15 s
there is still a current owing through the circuit.
Figure 9: Electric eld due to capacitor plates at point P.
(g)
The net electric eld at point P is in the direction of conventional current.
Figure 10: Net electric eld at point P.