SME
Grade 11 Science · Problem type
Speeding a charge through a voltage
Use this for “An electron starts at rest and crosses a 100 V potential difference. How much energy does it gain, in eV and in joules, and how fast does it go?”
Diagram
K = qΔV = 1 × 100 = 100 eV
K
=
1.6 ×
10
-17
J
v
=
√
2K/m
v
=
√
2 × 1.6 ×
10
-17
/9.11 ×
10
-31
v
=
5.93 ×
10
6
m/s
The electron is pulled from the − plate toward the +, speeding up all the way.
This is an example. Type your own values to start a new problem.
K (eV)
=
q (e)
×
ΔV (V)
m
Mass
example · kg
v
Speed
calculated · m/s
Formulas
K
=
q
×
ΔV
100 = 1 × 100
v
=
√
2 ×
K
× 1.602 ×
10
-19
/
m
5.93 ×
10
6
=
√(2
×
100
×
1.602 ×
10
-19
/(9.109
×
10
-31
))
Assumptions
•
It starts at rest and only the electric force does work on it: K = qΔV.
•
1 eV is the energy one electron charge gains through 1 V: 1 eV = 1.602 × 10⁻¹⁹ J.
•
Electron 9.109 × 10⁻³¹ kg, proton 1.673 × 10⁻²⁷ kg; speeds past a tenth of light’s need relativity.
Step-by-step
WE KNOW
ΔV = 100 V, q = 1 e, m = 9.109 × 10⁻³¹ kg
FIND
energy gained (K), speed (v)
Step 1 · Find energy gained (K)
Use K = q × ΔV
Each electron charge gains 1 eV for every volt it crosses.
K
=
1
×
100
K
=
100
eV
Step 2 · Find speed (v)
Use v = √(2 × K × 1.602 × 10⁻¹⁹/m)
Change
eV
to
joules,
then
solve
K
=
1
2
mv²
for
v
.
v
=
√(2
×
100
×
1.602 ×
10
-19
/(9.109
×
10
-31
))
K
=
100
eV
×
1.602 ×
10
-19
=
1.602 ×
10
-17
J
v
=
√(2
×
1.602 ×
10
-17
/(9.109
×
10
-31
))
v
=
5.93 ×
10
6
m
/s
Check
Put the numbers back in. Both sides should match.
100 = 1 × 100 ✓
5.93 ×
10
6
≈
√(2
×
100
×
1.602 ×
10
-19
/(9.109
×
10
-31
))
✓
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Charge and energy on a capacitor
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