Discussion Question 7A
P212, Week 7
RC Circuits
The circuit shown initially has the capacitor uncharged, and the switch connected to neither
terminal. At time t = 0, the switch is thrown to position a.
C
a
E
b
R1
R2
E = 12 V
C = 5 F
R1 = 3 
R2 = 6 
(a) At t = 0+, immediately after the switch is thrown to position a, what are the currents I1
and I2 across the two resistors?
What does the uncharged capacitor look like to the rest of the circuit at time 0? Does it offer
any resistance to the flow of charge? (Why or why not?)
At t = 0, the capacitor acts like wire w/ no resistance to current flow. Hence the battery is effectively
hooked to R2  6  I   / R2  12V / 6  2 A
(b) After a very long time, what is the instantaneous power P dissipated in the circuit?
After a very long time, what will have happened to the capacitor? Now what will it look like
to the rest of the circuit?
After a long time the capacitor acts like an open circuit and the battery current
flows through R1  R2 . Hence I =  /( R1  R2 .) = 12/9 =4/3 A and the power is
P   I  12V    4 / 3 A   16 W
(c) After a very long time, what is the Q charge on the capacitor?
To determine Q, you need the voltage across the capacitor ...
From part (b), we know after a long time I=4/3 A. The voltage drop across the capacitor
is the same as the voltage drop across R1  Vcap =IR1 =4/3 A  3  4V .
Q  C Vcap =  5 F  4V  =20C
Next, after a very long time T, the switch is thrown to position b.
C
a
E
b
R1
R2
(d) What is the time constant  that describes the discharging of the capacitor?
We have a nice formula available for time constants:  = RC. But the R in the formula refers
to the total resistance through which the capacitor discharges. Redrawing your circuit might
help you to determine this R .
When the switch is thrown to b, the capacitor discharges through R 1 and R 2 in parallel.
R equiv  R1 R 2 /  R1 + R 2   3(6) /(3  6)  2 ;   CRequiv   5 F  2   10 s
(e) Write down an equation for the time dependence of the charge on the capacitor, for
times t > T. Your answer for Q(t) should depend only on the known quantities E, R1, R2, C,
and T.
You know the general form for the time dependence of a discharging capacitor. All you have
to do is fix the constants in this expression to match the charge at t = T and at t = ∞.
We know that Q will involve a exp(-t/  factor added to a possible constant. The boundary
conditions are at t  T  , Q  20 C and fades to zero at infinity.
 t T 
The form that does this is Q  20 C exp  

 10 s 
(f) What is the charge Q20 on the capacitor 20 sec after time T?
 t T 
 20 s 
2
Q  20C exp  
  20C exp  
  20e  2.71  C
 10 s 
 10 s 
(g) What is the current through R 2 20 sec after time T ?
The voltage across R 2 is the Vcap  Q20 / C  2.71 C / 5 F  0.542 V
Vcap  I 2 R2  I 2   0.542 V  /  6   0.0903 A
2
Discussion Question 7B
P212, Week 7
Lorentz Force on Moving Charges
[FYI: the numbers shown in these solutions differ in a few places from the numbers in the
Discussion Problem Book.] The figure below depicts a region of space (shaded) where there is a
constant magnetic field, B = 1.2 T, directed along the negative z axis (into the page). The
magnetic field region extends infinitely far in the positive and negative y directions, but is
constrained to the region between x = 0 and x = 1 m. A proton travels along the x axis toward
this region, with initial speed v0.
y
proton:
m = 1.67 x 10-27 kg
q = e = 1.6 x 10-19 C
v0 = 2 x 108 m/s
B = 1.2 T
x
q, m
1m
(a) What is the radius of curvature of the proton's path after it enters the magnetic field
region?
To describe the particle's circular path, match the Lorentz force exerted by the B field to the
centripetal force.
The Lorentz force provides the centripetal force to allow the proton to accelerate about a circle
r
r r
Using F = ma in a direction perpendicular to the proton velocity: F = qv × B → F⊥ = qvB
v2
mv (1.67e − 27 kg )( 2e8 m / s )
F⊥ = qvB = ma = m → r =
=
= 1.74 m
r
qB
(1.e-19 C )(1.2Τ )
(b) At what x and y positions will the proton leave the magnetic field region?
Sketch the proton's path on the figure, and use your knowledge of the radius of curvature.
From figure x = 1 m and R sinθ = 1 → θ = sin −1 (1/1.74) = 0.612 rad.
y = R - R cosθ = 1.74 − 1.74cos ( 0.612 rad ) = 0.316m
R=1.74 m
R cosθ
θ
1m
x
y
proton:
m = 1.67 x 10-27 kg
q = e = 1.6 x 10-19 C
v0 = 2 x 108 m/s
B = 1.2 T
x
q, m
1m
(c) What is the magnitude v of the proton’s speed as it exits the magnetic field?
If the speed of the proton has changed, what has happened to its kinetic energy? Can the
Lorentz force cause such an effect?
Because the Lorentz force is always ⊥ to the proton path, it does
no work ⇒ no change in speed thus v = v0 = 2 × 108 m/s
(d) Suppose the proton had an initial component of velocity in the +z direction of
vz = 1 x 108 m/s, and the same initial speed v0 as before in the xy plane. What then would be
its final velocity v as it exits the magnetic field? This time, express your answer as a vector
with three components.
Analyze this situation by components. What effect does the
Lorentz force have on the particle's motion in the z direction?
What about in the xy-plane?
R=1.74 m
R cosθ
θ
The Lorentz force is in the x-y plane hence no z acceleration.
Thus vz is unchanged vz =1 × 108 m/s. From diagram you can
vx = v0 cosθ = 1 × 108 cos(0.612) = 1.63 × 108 m/s
1m
x
v0
θ
vx
v y = v0 sin θ = 1 × 108 sin(0.612) = 1.15 × 108 m/s
(e) Finally, a constant electric field E is added in the shaded region. The effect of this field is
that all charged particles launched with initial velocity v0 in the x direction travel straight through
the field region without being deflected at all! What is the magnitude and direction of E?
In order that the proton is undeflected, we need the electrical force to cancel the centripetal
force due to the magnetic force. The magnetic force is initially along the +yˆ axis. We thus
want the electric force to be along the -yˆ axis. To balance the forces we want:
r
FE = qE = FB = qBz v0 ⇒ E = Bz v0 = (1.2 T ) ( 2 × 108 m/s ) → E = −2.4 × 108 yˆ N/C
2
vy
Discussion Question 7C
P212, Week 7
RC Circuits
C
The circuit shown initially has the capacitor
uncharged, and the switch open. At time t = 0, the
switch is thrown. Write all answers in terms of E ,
R, and C as needed.
R
E
R
I
(a) At t  0 , immediately after the switch is
thrown what is the current I (0 ) ?
Consider applying KVL around the loop that includes the battery and the capacitor.
- Vcap  IR  0. At t=0+ Qcap =0  Vcap  0    IR  0  I (0 ) 
/R
(b) After a very long time, what is the current I  ?
When the capacitor is fully charged it acts like an open circuit. Hence all current
flows through R1  R2 . The current is then I =  / 2 R .
(c) After a very long time, what is the charge on the capacitor Q ?
Consider the KVL that just consists of the capacitor and the one resistor. I  R  Vcap  0
 I  R - Q / C  0 


R - Q / C  0  Q  C
2R
2
(d) Assume Q(t )  Q (1  exp(  t  gives the charge on the capacitor as a function of time. Use
 dQ 
 dQ 


 dt    I  0  and your answers to (a) and (c) to compute  . Why does  dt    I  0  ?
0
0
What does your  imply for the effective resistance that you use in   Reffective C ?
Initially there is no current flow through the resistor in parallel with the capacitor
Hence I(0) flows entirely through the capacitor. The rate of change of charge
dQ 
 dQ 
d

0   I  0  .    Q  1  e   t     Q

dt
 dt  0
 dt
0
2

RC
R
 dQ 
 C 

 dt    Q    2   I  0   R    RC    2  Reffective  2
 0


Essentially the two resistors act as if they were in parallel during the charging operation.
We would get the same time constant if we discharged the capacitor by shorting the battery/
is the current hence
(e) Use KVL for a loop that includes the capacitor and the battery to compute I (t ) . Check
that your answers are consistent with your answers to parts (a) and (b).
1
1 
1  C 
Q
Q
 2t    
 IR  0  I       I    

1
exp

 
C
R
C
R 
C  2 
 RC    
  1 1
 2t     
 2t  
 I  1    exp  
1  exp  
  


R  2 2
 RC    2 R 
 RC  
  Vcap  IR  0   

 2t 
At t  0 ,1  exp  
  2  I  which agrees with (a)
R
 RC 

 2t 
As t  ,1  exp  
which agrees with (b)
 1 I 
2R
 RC 
2