SME
Electromagnetic Theory
Electrostatics
Use this for “A thin ring of radius 15 cm carries −6 nC spread evenly. Find the potential and the field on its axis 20 cm from its center.”
Diagram
V
=
kQ
÷
√
z
2
+
R
2
V
=
8.99 ×
10
9
×
1 ×
10
-8
0.5
=
180
V
On
the
axis
E
=
kQz
÷
(
z
2
+
R
2
)
3
2
=
8.99 ×
10
9
×
1 ×
10
-8
×
0.4
0.5
3
=
288
N/C
The sideways parts of dE from opposite pieces cancel; the parts along the axis add.
This is an example. Type your own values to start a new problem.
V
=
k
Q (
)
(
z (
)
)
2
+
(
R (
)
)
2
=
V (
)
s
Distance from each piece to P
worked out (not typed)
Ez
Field along the axis
calculated
Formulas
s
=
√
z
2
+
R
2
0.5
=
√
0.4
2
+
0.3
2
V
=
8.99 ×
10
9
×
Q
s
+179.8
=
8.99 ×
10
9
×
1 ×
10
-8
0.5
Ez
=
8.99 ×
10
9
×
Q
×
z
s
3
+287.7
=
8.99 ×
10
9
×
(1
×
10
-8
)
×
0.4
0.5
3
Formulas are worked in C. The step-by-step shows the conversions.
Assumptions
•
The charge Q is spread evenly round a thin ring of radius R; P is on the axis, z from the center; k = 8.99 × 10⁹ N·m²/C².
•
V first: V = ∫ k dq ÷ s, and every piece is the same s = √(z² + R²) away, so V = kQ ÷ s.
•
E from the slope of V: Ez = −dV/dz = kQz ÷ s³. On the axis the sideways parts of dE cancel, so E points along the axis (away from the ring when Q > 0).
Step-by-step
WE KNOW
Q = 10 nC, R = 0.3 m, z = 0.4 m
FIND
distance from each piece to P (s), potential at P (V), 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).
Q = 10 nC = 1 × 10⁻⁸ C (1 C = 10⁹ nC)
Step 2 · Find distance from each piece to P (s)
Use s = √(z² + R²)
Pythagoras:
z
along
the
axis
and
R
out
to
the
ring
are
at
right
angles.
s
=
√
0.4
2
+
0.3
2
s
=
√
0.16 + 0.09
s
=
√
0.25
s
=
0.5
m
Step 3 · Find potential at P (V)
Use V = 8.99 × 10⁹ × Q ÷ s
V
=
∫
k
dq
s
=
(
k
s
)
∫
dq
=
kQ
s
:
every
piece
dq
is
the
same
s
away,
so
the
pieces
add
to
kQ
s
.
V
=
8.99 ×
10
9
×
10
-8
0.5
V
=
89.9
0.5
V
=
+179.8
V
Step 4 · Find field along the axis (Ez)
Use Ez = 8.99 × 10⁹ × Q × z ÷ s³
Ez
=
−dV/dz
=
−kQ
×
(−
1
2
)
(
z
2
+
R
2
)
−
3
2
×
2
z
=
kQz
s
3
:
minus
the
slope
of
V
along
z
,
the
chain
rule
bringing
down
z
.
Ez
=
8.99 ×
10
9
×
10
-8
×
0.4
0.5
3
Ez
=
(8.99
×
10
9
)
×
10
-8
×
0.4
0.125
Ez
=
89.9
×
0.4
0.125
Ez
=
35.96
0.125
Ez = +287.7 N/C
Check
Put the numbers back in. Both sides should match.
0.5
=
√
0.4
2
+
0.3
2
✓
+179.8
=
8.99 ×
10
9
×
10
-8
0.5
✓
+287.7
=
8.99 ×
10
9
×
10
-8
×
0.4
0.5
3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
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.”