SME
Electromagnetic Theory · Problem type
Displacement current: B between charging plates
Use this for “Round capacitor plates of radius 4 cm charge with a current of 1.5 A. Find the displacement current and B 2.5 cm from the axis, between the plates.”
Diagram
dE/dt
=
I
ε₀πR²
dE/dt
=
2.88 ×
10
13
V/(m·s)
Displacement
current
inside
r:
I
(
r
R
)
2
=
2
×
(
0.03
0.05
)
2
=
0.72
A
Round
the
circle,
B
=
μ₀(current inside)
2πr
=
1.26 ×
10
-6
×
0.72
2π × 0.03
=
4.8 ×
10
-6
T
This is an example. Type your own values to start a new problem.
B (
)
=
μ₀
I (
)
r (
)
2π
(
R (
)
)
2
dE/dt
Rate E grows
calculated · V/(m·s)
Id
Displacement current inside r
calculated
Formulas
dE/dt
=
I
8.85 ×
10
-12
× π ×
R
2
2.8774 ×
10
13
=
2
8.85 ×
10
-12
× π ×
0.05
2
Id
=
I
×
(
r
R
)
2
0.72
=
2
×
(
0.03
0.05
)
2
B
=
4π
×
10
-7
×
Id
2π ×
r
4.8 ×
10
-6
=
4π
×
10
-7
×
0.72
2π × 0.03
Formulas are worked in m, T. The step-by-step shows the conversions.
Assumptions
•
Round parallel plates close together: between them E is even and square to the plates, E = σ ÷ ε₀, with ε₀ = 8.85 × 10⁻¹² F/m.
•
Ampère–Maxwell: round a loop, B × length = μ₀(Ithrough + ε₀ dΦE/dt); no charge crosses the gap, so only the displacement current counts.
•
A circle of radius r ≤ R takes in the share (r/R)² of the plates’ area, so of the displacement current; μ₀ = 4π × 10⁻⁷ T·m/A.
Step-by-step
WE KNOW
I = 2 A, R = 5 cm, r = 3 cm
FIND
rate E grows (dE/dt), displacement current inside r (Id), field at r (B)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m, T).
R = 5 cm = 0.05 m (1 m = 100 cm)
r = 3 cm = 0.03 m (1 m = 100 cm)
Step 2 · Find rate E grows (dE/dt)
Use dE/dt = I ÷ (8.85 × 10⁻¹² × π × R²)
The
charge
on
the
plates
grows
at
I
,
so
E
=
Q
ε₀πR²
grows
at
I
ε₀πR²
.
dE/dt
=
2
8.85 ×
10
-12
× π ×
0.05
2
dE/dt
=
2
(8.85 ×
10
-12
) × 0.0025 × π
dE/dt
=
2
(2.2125 ×
10
-14
) × π
dE/dt
=
2
6.951 ×
10
-14
dE/dt
=
2.8774 ×
10
13
V/(m·s)
Step 3 · Find displacement current inside r (Id)
Use Id = I × (r ÷ R)²
ε₀
times
the
rate
of
the
flux
through
the
circle,
ε₀πr²
dE/dt:
the
share
(
r
R
)
2
of
I
.
Id
=
2
×
(
0.03
0.05
)
2
Id
=
2
×
0.6
2
Id = 2 × 0.36
Id = 0.72 A
Step 4 · Find field at r (B)
Use B = 4π × 10⁻⁷ × Id ÷ (2π × r)
Ampère–Maxwell
round
the
circle:
B
×
2πr
=
μ₀Id,
as
for
a
wire
carrying
Id.
B
=
4π
×
10
-7
×
0.72
2π × 0.03
B
=
4.8 ×
10
-6
T
Step 5 · Convert the answers
Change each answer back to the unit you chose.
B = 4.8 × 10⁻⁶ T = 4.8 μT (1 T = 10⁶ μT)
Check
Put the numbers back in. Both sides should match.
2.8774 ×
10
13
=
2
8.85 ×
10
-12
× π ×
0.05
2
✓
0.72
=
2
×
(
0.03
0.05
)
2
✓
4.8 ×
10
-6
=
4π
×
10
-7
×
0.72
2π × 0.03
✓
Topic check
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