Discussion Session for PHY2049
Physics with Calculus 2 - Electromagnetism
Spring 2012
Week # 12
Problem 1
In the figure, a parallel plate capacitor has square plates of edge length L= 1.0 m. A current
~ between the plates,
of 2.0 A charges the capacitor, producing a uniform electric field E
~ perpendicular to the plates. (a) What is the displacement current id through the
with E
region between the plates? (b)What is dE/dt in this region? (c) What is the displacement
H
~ around
~ · ds
current encircled by the square dashed path of edge d=0.5m? (d) What is B
this square dashed path?
Solution
a) The charge in a capacitor is related to the voltage across its plates by q = CV . In a
parallel plate capacitor we have that C =
0 A
s
where s is the separation between the plates.
Also, since the electric field inside the plates is almost uniform we have that V = Es, hence
q = CV = CEs =
0 A
Es = 0 AE
s
Taking the time derivative of the expression above we obtain
dq
dE
= 0 A
dt
dt
dE
I = 0 A
dt
The displacement current is then
id = 0
dΦE
d(E · A)
dE
I
= 0
= 0 A
= 0 A
=I
dt
dt
dt
0 A
Thus, the lesson is that the displacement current is ALWAYS equal to the current that is
charging the capacitor. The “always” here means “at all times”. Therefore, for a current
I = 2A the displacement current is id = 2A.
b) We’ve already done the work for this question. This is
dE
I
2
=
=
= 2.3 × 1011 V/m·s
dt
0 A
8.85 × 10−12 (1)2
1
c) Since the ratio id /Area is constant1 we have
idenc =
d2
0.52
Aenc
id = 2 id = 2 2 = 0.5A
Atot
L
1
(1)
d) Since there’s no conduction current through the plate, Ampere-Maxwell law reads
I
~ = µ0 i d
~ · ds
B
(2)
Note that usually the right hand side is µ0 ienc + µ0 id , but here there’s no real current
enclosed by the loop, i.e. µ0 ienc = 0. Thus
I
~ = µ0 id = 4π × 10−7 × 0.5 = 6.3 × 10−7 T·m
~ · ds
B
1
The electric field inside the capacitor is uniform
2
(3)