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Grade 11 Science · Problem type
The field between charged plates
Use this for “Two plates 3 mm apart have 12 V across them. What is the field between them, and the force on an electron there?”
Diagram
Field: E = V/d = 12/0.003 = 4,000 V/m
Force: F = qE = (−1.6 × 10⁻¹⁹) × 4,000 = −6.41 × 10⁻¹⁶ N
A − charge is pushed against the field, toward the + plate.
Between the plates the field is the same everywhere, so the force is the same wherever the charge is.
This is an example. Type your own values to start a new problem.
E (V/m)
=
V (V)
d (m)
q
Charge
example · C
F
Force on the charge
calculated · N
Formulas
E
=
V
d
4,000 = 12/0.003
F
=
q
×
E
−6.41 ×
10
-16
=
−1.602 ×
10
-19
×
4,000
Assumptions
•
Between large parallel plates the field is uniform: E = V/d, from the + plate to the − plate.
•
A charge there feels F = qE: a + charge along the field, a − charge against it.
•
An electron’s charge is −1.602 × 10⁻¹⁹ C; a proton’s is +1.602 × 10⁻¹⁹ C.
•
Air between the plates sparks over above about 3 × 10⁶ V/m.
Step-by-step
WE KNOW
V = 12 V, d = 0.003 m, q = −1.602 × 10⁻¹⁹ C
FIND
field between the plates (E), force on the charge (F)
Step 1 · Find field between the plates (E)
Use E = V/d
The voltage for each meter of the gap.
E
=
12/0.003
E
=
4,000
V/m
Step 2 · Find force on the charge (F)
Use F = q × E
Charge times field; its sign gives the direction.
F
=
−1.602 ×
10
-19
×
4,000
F
=
−6.41 ×
10
-16
N
Check
Put the numbers back in. Both sides should match.
4,000 = 12/0.003 ✓
−6.41 ×
10
-16
≈
−1.602 ×
10
-19
×
4,000
✓
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Charged by friction, conduction or induction?
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