SME
Engineering Electromagnetics · Problem type
Displacement current in a capacitor
Use this for “A 1 μF capacitor’s voltage rises at 1000 V/s. Find the displacement current between its plates.”
Diagram
Q = CV = 1 × 10⁻⁶ × 5 = 5 × 10⁻⁶ C
U
=
1
2
CV²
=
1
2
×
1 ×
10
-6
F
×
5
2
U
=
1.25 ×
10
-5
J
Twice the voltage: twice the charge and four times the energy.
Rate of voltage rise dV/dt = 1,000 V/s
Displacement current id = 1 mA
Time since 0 V t = 5 ms
This is an example. Type your own values to start a new problem.
id (mA)
=
C (μF)
×
dV/dt (V/s)
t
Time since 0 V
example
V
Voltage at t
calculated
Q
Charge at t
calculated
Formulas
id
=
C
×
(dV/dt)
0.001
=
1 ×
10
-6
×
1,000
V
=
(dV/dt)
×
t
5 = 1,000 × 0.005
Q
=
C
×
V
5 ×
10
-6
=
1 ×
10
-6
×
5
Formulas are worked in F, A, s, C. The step-by-step shows the conversions.
Assumptions
•
The voltage across the plates rises at a steady rate dV/dt, in volts per second, from 0 V.
•
No charge crosses the gap, but E grows there: the displacement current is id = ε₀A dE/dt = C dV/dt.
•
id equals the conduction current in the wires, so Ampère’s law holds through the gap too.
•
After a time t the voltage is V = (dV/dt)t, and the plates hold Q = CV.
Step-by-step
WE KNOW
C = 1 μF, dV/dt = 1,000 V/s, t = 5 ms
FIND
displacement current (id), voltage at t (V), charge at t (Q)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to F, A, s, C).
C = 1 μF = 1 × 10⁻⁶ F (1 F = 10⁶ μF)
t = 5 ms = 0.005 s (1 s = 1000 ms)
Step 2 · Find displacement current (id)
Use id = C × (dV/dt)
Displacement
current:
ε₀A
dE/dt
across
the
gap
is
C
dV/dt,
the
same
as
the
wires’
current.
id
=
10
-6
×
1,000
id = 0.001 A
Step 3 · Find voltage at t (V)
Use V = (dV/dt) × t
A steady rise from 0 V: the rate times the time.
V
=
1,000
×
0.005
V
=
5
V
Step 4 · Find charge at t (Q)
Use Q = C × V
The
charge
on
the
plates
is
C
times
the
voltage
across
them.
Q
=
10
-6
×
5
Q
=
5 ×
10
-6
C
Step 5 · Convert the answers
Change each answer back to the unit you chose.
id = 0.001 A = 1 mA (1 A = 1000 mA)
Q = 5 × 10⁻⁶ C = 5 μC (1 C = 10⁶ μC)
Check
Put the numbers back in. Both sides should match.
0.001
=
10
-6
×
1,000
✓
5 = 1,000 × 0.005 ✓
5 ×
10
-6
=
10
-6
×
5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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