SME
Grade 11 Science · Problem type
A parallel-plate capacitor
Use this for “Two 0.01 m² plates sit 1 mm apart in air. What is the capacitance, what charge does 12 V put on it, and how strong is the field?”
Diagram
C = κε₀A/d
C
=
1
×
8.85 ×
10
-12
×
0.01/0.001
C = 88.5 pF
Q = CV = 88.5 × 12 = 1,060 pC
U
=
1
2
CV²
U
=
1
2
×
8.85 ×
10
-11
F
×
12
2
U
=
6.37 ×
10
-9
J
Twice the voltage: twice the charge and four times the energy.
Field between the plates: 12,000 V/m
This is an example. Type your own values to start a new problem.
C (pF)
=
κ
×
8.85
pF/m
×
A (m²)
d (mm)
V
Voltage
example
Q
Charge
calculated
E
Field between the plates
calculated · V/m
Formulas
C
=
κ
×
8.85
×
A
d
88.5 = 1 × 8.85 × 0.01/0.001
Q
=
C
×
V
1,062 = 88.5 × 12
E
=
V
d
12,000 = 12/0.001
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
The gap is small next to the plates, so the field between them is even: C = κε₀A/d.
•
ε₀ = 8.85 × 10⁻¹² F/m, which is 8.85 pF/m; κ = 1 for air, and an insulator between the plates raises C κ times.
•
Bigger plates or a smaller gap hold more charge at the same voltage.
Step-by-step
WE KNOW
κ = 1, A = 0.01 m², d = 1 mm, V = 12 V
FIND
capacitance (C), charge (Q), field between the plates (E)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
d = 1 mm = 0.001 m (1 m = 1000 mm)
Step 2 · Find capacitance (C)
Use C = κ × 8.85 × A/d
κ
times
ε₀
=
8.85
pF/m
times
the
area,
over
the
gap
in
meters.
C
=
1
×
8.85
×
0.01/0.001
C
=
0.0885/0.001
C
=
88.5
pF
Step 3 · Find charge (Q)
Use Q = C × V
Each
volt
puts
C
more
picocoulombs
on
each
plate.
Q
=
88.5
×
12
Q
=
1,062
pC
Step 4 · Find field between the plates (E)
Use E = V/d
The field is even between the plates: the voltage over the gap in meters.
E
=
12/0.001
E
=
12,000
V/m
Check
Put the numbers back in. Both sides should match.
88.5 = 1 × 8.85 × 0.01/0.001 ✓
1,062 = 88.5 × 12 ✓
12,000 = 12/0.001 ✓
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Charge and energy on a capacitor
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Circuits: Ohm's law, series and parallel
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