SME
University Physics II: Electricity & Magnetism · Problem type
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.”
Diagram
E
×
2πrℓ
=
λℓ
ε₀
,
so
E
=
2k
λ
r
=
2
×
8.99 ×
10
9
×
5 ×
10
-9
0.2
=
450
N/C
No flux leaves through the cylinder’s flat ends: E is square to the rod.
This is an example. Type your own values to start a new problem.
E (N/C)
=
2k
λ (
)
r (m)
Formulas
E
=
2
×
8.99 ×
10
9
×
λ
r
449.5
=
2
×
8.99 ×
10
9
×
5 ×
10
-9
0.2
Formulas are worked in C/m. The step-by-step shows the conversions.
Assumptions
•
The line is straight and much longer than r, so E points straight out from it and is the same all round a cylinder of radius r.
•
Gauss’s law on that cylinder, length ℓ: E × 2πrℓ = λℓ ÷ ε₀, and no flux leaves its flat ends.
•
So E = λ ÷ (2πε₀r) = 2kλ ÷ r, with k = 8.99 × 10⁹ N·m²/C²; a negative λ gives a negative E, pointing in.
Step-by-step
WE KNOW
λ = 5 nC/m, r = 0.2 m
FIND
electric field at r (E)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to C/m).
λ = 5 nC/m = 5 × 10⁻⁹ C/m (1 C/m = 10⁹ nC/m)
Step 2 · Find electric field at r (E)
Use E = 2 × 8.99 × 10⁹ × λ ÷ r
Gauss’s
law
on
a
cylinder
round
the
line:
λ
2πε₀
=
2kλ,
spread
over
the
distance
r
.
E
=
2
×
8.99 ×
10
9
×
5 ×
10
-9
0.2
E
=
(1.798
×
10
10
)
×
5 ×
10
-9
0.2
E
=
89.9
0.2
E
=
449.5
N/C
Check
Put the numbers back in. Both sides should match.
449.5
=
2
×
8.99 ×
10
9
×
5 ×
10
-9
0.2
✓
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Electric fields and Gauss's law
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Gauss’s law: the field of a charged sheet
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Related lessons
Electric fields and Gauss's law
Main lesson
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.”
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.”