SME
Engineering Electromagnetics · Problem type
Gauss’s law: E of a line charge
Use this for “Find E 0.1 m from a long line charge of 1 nC/m.”
Diagram
E
×
2πrℓ
=
λℓ
ε₀
,
so
E
=
λ
2πε₀r
=
1 ×
10
-9
2π × 8.854 ×
10
-12
× 0.1
=
180
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 (V/m)
=
λ (nC/m)
2πε₀
r (m)
Formulas
E
=
λ
2π × 8.854 ×
10
-12
×
r
+179.7548
=
1 ×
10
-9
2π × 8.854 ×
10
-12
× 0.1
Formulas are worked in C/m. The step-by-step shows the conversions.
Assumptions
•
A long straight line of even charge λ; r is measured square to the line.
•
Gauss’s law on a cylinder of radius r and length ℓ round the line: E × 2πrℓ = λℓ ÷ ε₀.
•
In air, ε₀ = 8.854 × 10⁻¹² F/m; a negative λ gives a negative E, pointing in toward the line.
Step-by-step
WE KNOW
λ = +1 nC/m, r = 0.1 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 = +1 nC/m = 1 × 10⁻⁹ C/m (1 C/m = 10⁹ nC/m)
Step 2 · Find electric field at r (E)
Use E = λ ÷ (2π × 8.854 × 10⁻¹² × r)
Gauss’s
law:
the
flux
λℓ
ε₀
leaves
through
the
cylinder’s
side,
of
area
2πrℓ.
E
=
10
-9
2π × 8.854 ×
10
-12
× 0.1
E
=
+179.7548
V/m
Check
Put the numbers back in. Both sides should match.
+179.7548
=
10
-9
2π × 8.854 ×
10
-12
× 0.1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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