SME
Grade 11 Science · Physics
Electric potential, voltage and capacitors
Use this for “What is the potential 0.5 m from a 4 μC charge, and the potential energy of a 2 μC charge placed there?”
Diagram
V = kq/r
V
=
8.99 ×
10
9
×
4 ×
10
-6
/0.5
V = 71,900 V
U
=
q₀V
=
2 ×
10
-6
×
71,900
U = 0.144 J
Half as far, twice the potential: V falls off as 1/r, not 1/r².
Moving a charge along a dashed circle takes no work: V is the same all the way round.
This is an example. Type your own values to start a new problem.
V (V)
=
8.99
×
10⁹
×
q (μC)
r (m)
q₀
Second charge
example
U
Potential energy
calculated · J
Formulas
V
=
8.99 ×
10
9
×
q
×
10
-6
/
r
7.19 ×
10
4
=
8.99 ×
10
9
×
4 ×
10
-6
/0.5
U
=
q₀
×
10
-6
×
V
0.144
=
2 ×
10
-6
×
(7.19
×
10
4
)
Assumptions
•
Potential V is energy per coulomb, counted from far away, where V = 0: V = kq/r.
•
V takes the charge’s sign and has no direction, so potentials add as plain numbers.
•
A second charge there has potential energy U = q₀V: like charges give U > 0, the work done to push them together.
Step-by-step
WE KNOW
q = 4 μC, r = 0.5 m, q₀ = 2 μC
FIND
potential (V), potential energy (U)
Step 1 · Find potential (V)
Use V = 8.99 × 10⁹ × q × 10⁻⁶/r
k
=
8.99 ×
10
9
N·
m
2
/
C
2
times
the
charge
in
coulombs,
over
r
(not
r
2
).
V
=
8.99 ×
10
9
×
4 ×
10
-6
/0.5
q
=
4
μC
=
4 ×
10
-6
C
V
=
7.19 ×
10
4
V
Step 2 · Find potential energy (U)
Use U = q₀ × 10⁻⁶ × V
The potential is joules per coulomb: times the second charge in coulombs.
U
=
2 ×
10
-6
×
(7.19
×
10
4
)
U
=
0.144
J
Check
Put the numbers back in. Both sides should match.
7.19 ×
10
4
≈
8.99 ×
10
9
×
4 ×
10
-6
/0.5
✓
0.144
≈
2 ×
10
-6
×
(7.19
×
10
4
)
✓
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