SME
Grade 11 Science · Problem type
Charge and energy on a capacitor
Use this for “A 470 μF capacitor is charged to 9 V. How much charge does it hold, and how much energy does it store?”
Diagram
Q = CV = 470 × 9 = 4,230 μC
U
=
1
2
CV²
=
1
2
×
4.7 ×
10
-4
F
×
9
2
U = 0.019 J
Twice the voltage: twice the charge and four times the energy.
This is an example. Type your own values to start a new problem.
Q (μC)
=
C (μF)
×
V (V)
U
Stored energy
calculated · J
Formulas
Q
=
C
×
V
4,230 = 470 × 9
U
=
1
2
×
C
×
10
-6
×
V
2
0.0190
=
1
2
×
470 ×
10
-6
×
9
2
Assumptions
•
Q is the charge on each plate, +Q on one and −Q on the other: Q = CV.
•
The energy is stored in the field between the plates: U = ½CV², so twice the voltage stores four times the energy.
•
1 μF = 10⁻⁶ F, so μF × V gives μC.
Step-by-step
WE KNOW
C = 470 μF, V = 9 V
FIND
charge (Q), stored energy (U)
Step 1 · Find charge (Q)
Use Q = C × V
Each
volt
puts
C
more
microcoulombs
on
each
plate.
Q
=
470
×
9
Q
=
4,230
μC
Step 2 · Find stored energy (U)
Use U = ½ × C × 10⁻⁶ × V²
Half the capacitance in farads times the voltage squared.
U
=
1
2
×
470 ×
10
-6
×
9
2
U
=
1
2
×
4.7 ×
10
-4
×
81
U
=
0.0190
J
Check
Put the numbers back in. Both sides should match.
4,230 = 470 × 9 ✓
0.0190
≈
1
2
×
470 ×
10
-6
×
9
2
✓
Topic check
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