SME
Electromagnetic Theory · Problem type
The field of a charged disk, on its axis
Use this for “A disk of radius 10 cm carries 2 μC/m². Find the field on its axis 5 cm from its center, and compare it with a wide sheet’s field.”
Diagram
The sheet’s field 2πkσ = 56,500 N/C
Adding
rings
from
the
center
to
R:
E
=
2πkσ(1
−
z
÷
√
z
2
+
R
2
)
=
56,500
×
(1
−
0.4
0.5
)
=
11,300
N/C
Close to the disk E nears the sheet’s field; far away the disk acts as a point charge.
This is an example. Type your own values to start a new problem.
E
sheet
=
2πk
σ (
)
=
Eₛₕₑₑₜ (
)
E
z
=
Eₛₕₑₑₜ
1
−
z
z
2
+
R
2
=
Ez (
)
Formulas
Eₛₕₑₑₜ
=
2
×
π
×
8.99 ×
10
9
×
σ
+56,486
=
2
×
π
×
8.99 ×
10
9
×
(1
×
10
-6
)
Ez
=
Eₛₕₑₑₜ
×
(1
−
z
÷
√
z
2
+
R
2
)
+11,297
=
56,486
×
(1
−
0.4
÷
√
0.4
2
+
0.3
2
)
Formulas are worked in C/m². The step-by-step shows the conversions.
Assumptions
•
The charge is spread evenly over a thin flat disk, σ per square meter; P is on the axis, z from the center; k = 8.99 × 10⁹ N·m²/C².
•
The disk is rings of radius r, each with dq = σ × 2πr dr; on the axis each ring’s sideways parts cancel, and the rings from 0 to R add.
•
Close in (z much less than R) E nears the sheet’s 2πkσ = σ ÷ (2ε₀); far away the disk acts as a point charge.
Step-by-step
WE KNOW
σ = 1 μC/m², R = 0.3 m, z = 0.4 m
FIND
wide sheet’s field (Eₛₕₑₑₜ), field along the axis (Ez)
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 = 1 μC/m² = 1 × 10⁻⁶ C/m² (1 C/m² = 10⁶ μC/m²)
Step 2 · Find wide sheet’s field (Eₛₕₑₑₜ)
Use Eₛₕₑₑₜ = 2 × π × 8.99 × 10⁹ × σ
The
field
of
a
wide
sheet
with
the
same
σ
:
σ
2ε₀
,
which
is
2πkσ.
Eₛₕₑₑₜ
=
2
×
π
×
8.99 ×
10
9
×
10
-6
Eₛₕₑₑₜ
=
(1.798
×
10
10
)
×
10
-6
×
π
Eₛₕₑₑₜ = 17,980 × π
Eₛₕₑₑₜ = +56,486 N/C
Step 3 · Find field along the axis (Ez)
Use Ez = Eₛₕₑₑₜ × (1 − z ÷ √(z² + R²))
Ez
=
∫₀ᴿ
kσ2πrz
dr
÷
(
z
2
+
r
2
)
3
2
=
2πkσz[
1
z
−
1
÷
√
z
2
+
R
2
]:
the
rings
from
the
center
to
the
rim
added
up,
the
sheet’s
field
times
1
−
z
÷
√
z
2
+
R
2
.
Ez
=
56,486
×
(1
−
0.4
÷
√
0.4
2
+
0.3
2
)
Ez
=
56,486
×
(1
−
0.4
÷
√
0.16 + 0.09
)
Ez
=
56,486
×
(1
−
0.4
÷
√
0.25
)
Ez
=
56,486
×
(1
−
0.4
0.5
)
Ez = 56,486 × (1 − 0.8)
Ez = 56,486 × 0.2
Ez = +11,297 N/C
Check
Put the numbers back in. Both sides should match.
+56,486
=
2
×
π
×
8.99 ×
10
9
×
10
-6
✓
+11,297
=
56,486
×
(1
−
0.4
÷
√
0.4
2
+
0.3
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Electrostatics
Main lesson
The method of images: a charge above a grounded plane
Use this for “A 4 nC charge is held 2 cm above a large grounded metal plate. Find the force pulling it toward the plate, the induced charge per area just below it, and the total induced charge.”
The potential of an electric dipole, far away
Use this for “Charges of +3 nC and −3 nC sit 2 mm apart. Find the potential 5 cm from their center, 45° from the axis on the + side.”