SME
University Physics II: Electricity & Magnetism · Problem type
Gauss’s law: the field of a charged sheet
Use this for “A large flat sheet carries 4 nC/m². Find the field beside it, and between two sheets carrying +4 and −4 nC/m².”
Diagram
E
×
2A
=
σA
ε₀
,
so
E
=
σ
2ε₀
=
8.85 ×
10
-9
2 × 8.85 ×
10
-12
=
500
N/C
E is the same at every distance from the sheet.
Between
opposite
sheets
the
fields
add:
E
=
σ
ε₀
=
1,000
N/C,
and
0
outside
This is an example. Type your own values to start a new problem.
E (N/C)
=
σ (
)
2ε₀
Eb
Field between two opposite sheets
calculated · N/C
Formulas
E
=
σ
2 × 8.85 ×
10
-12
500
=
8.85 ×
10
-9
2 × 8.85 ×
10
-12
Eb
=
σ
8.85 ×
10
-12
1,000
=
8.85 ×
10
-9
8.85 ×
10
-12
Formulas are worked in C/m². The step-by-step shows the conversions.
Assumptions
•
The sheet is flat and much wider than the distance to it, so E points straight out from it, the same on both sides.
•
Gauss’s law on a pillbox through the sheet, faces of area A: E × 2A = σA ÷ ε₀, so E = σ ÷ (2ε₀), with ε₀ = 8.85 × 10⁻¹² F/m.
•
E does not depend on the distance from the sheet, as long as the sheet still looks endless from there.
•
Two sheets with +σ and −σ: their fields add between them, Eb = σ ÷ ε₀, and cancel outside.
Step-by-step
WE KNOW
σ = 8.85 nC/m²
FIND
field of one sheet (E), field between two opposite sheets (Eb)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to C/m²).
σ = 8.85 nC/m² = 8.85 × 10⁻⁹ C/m² (1 C/m² = 10⁹ nC/m²)
Step 2 · Find field of one sheet (E)
Use E = σ ÷ (2 × 8.85 × 10⁻¹²)
Gauss’s
law
on
the
pillbox:
the
flux
leaves
through
both
faces,
so
2EA
=
σA
ε₀
;
a
negative
σ
gives
E
pointing
in.
E
=
8.85 ×
10
-9
2 × 8.85 ×
10
-12
E
=
8.85 ×
10
-9
1.77 ×
10
-11
E
=
500
N/C
Step 3 · Find field between two opposite sheets (Eb)
Use Eb = σ ÷ (8.85 × 10⁻¹²)
Between
the
+
σ
and
−
σ
sheets
both
fields
point
the
same
way,
from
+
to
−,
so
they
add:
twice
one
sheet’s
field.
Eb
=
8.85 ×
10
-9
8.85 ×
10
-12
Eb = 1,000 N/C
Check
Put the numbers back in. Both sides should match.
500
=
8.85 ×
10
-9
2 × 8.85 ×
10
-12
✓
1,000
=
8.85 ×
10
-9
8.85 ×
10
-12
✓
← Previous lesson
Gauss’s law: the field of a long line of charge
Next lesson →
Superposition: the field of two charges on a line
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Electric fields and Gauss's law
Main lesson
Gauss’s law: the field of a long line of charge
Use this for “A long wire carries 3 nC/m. Find the field 0.15 m from it, and how far out the field drops to 100 N/C.”
Superposition: the field of two charges on a line
Use this for “A +3 μC charge sits at the origin and a −2 μC charge at x = 0.4 m. Find the field at x = 0.6 m, beyond both.”