SME
University Physics II: Electricity & Magnetism · Problem type
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.”
Diagram
Field lines leave + charges and end on − charges; where they crowd, the field is strong.
At
x
=
0.1
m:
E₁
=
kq₁/
x
2
=
3,600,000
N/C
and
E₂
=
kq₂/
(x − r)
2
=
−225,000
N/C,
each
+
pointing
toward
q₂’s
side.
E = E₁ + E₂
E = 3,600,000 + (−225,000)
E = 3,370,000 N/C
Each charge’s field points away from it if +, toward it if −; the fields add as arrows.
This is an example. Type your own values to start a new problem.
q₁
Charge at the origin
example · μC
q₂
Second charge
example · μC
x₂
Position of q₂
example · m
x
Position of the point
example · m
E₁
Field from q₁ at the point
calculated · N/C
E₂
Field from q₂ at the point
calculated · N/C
E
Net field at the point
calculated · N/C
Formulas
E₁
=
8.99 ×
10
9
×
q₁
×
10
-6
×
x
÷
|
x
|³
+3,596,000
=
8.99 ×
10
9
×
4 ×
10
-6
×
0.1
÷
|0.1|³
E₂
=
8.99 ×
10
9
×
q₂
×
10
-6
×
(
x
−
x₂
)
÷
|
x
−
x₂
|³
−224,750
=
8.99 ×
10
9
×
1 ×
10
-6
×
(0.1
−
0.3)
÷
|0.1
−
0.3|³
E
=
E₁
+
E₂
+3,371,250 = 3,596,000 + (−224,750)
Assumptions
•
The charges and the point lie on the x-axis: q₁ at 0, q₂ at x₂; a field is + when it points toward +x.
•
Each charge’s field has size kq ÷ d², d its distance to the point, with k = 8.99 × 10⁹ N·m²/C²; writing kq(x − x₀) ÷ |x − x₀|³ gives its sign too.
•
A field points away from a + charge and toward a − charge, so its sign depends on which side the point is.
•
Superposition: the net field is the sum of the two fields, signs and all.
Step-by-step
WE KNOW
q₁ = +4 μC, q₂ = +1 μC, x₂ = 0.3 m, x = 0.1 m
FIND
field from q₁ at the point (E₁), field from q₂ at the point (E₂), net field at the point (E)
Step 1 · Find field from q₁ at the point (E₁)
Use E₁ = 8.99 × 10⁹ × q₁ × 10⁻⁶ × x ÷ |x|³
The
point
is
on
q₁
’s
right:
a
+
q₁
pushes
its
field
toward
+
x
,
so
E₁
takes
q₁
’s
sign.
E₁
=
8.99 ×
10
9
×
q₁
×
10
-6
x
2
E₁
=
8.99 ×
10
9
×
4 ×
10
-6
0.1
2
E₁
=
(8.99
×
10
9
)
×
4 ×
10
-6
0.01
E₁
=
35,960
0.01
E₁
=
+3,596,000
N/C
Step 2 · Find field from q₂ at the point (E₂)
Use E₂ = 8.99 × 10⁹ × q₂ × 10⁻⁶ × (x − x₂) ÷ |x − x₂|³
The
point
is
on
q₂
’s
left,
x₂
−
x
away:
a
+
q₂
pushes
its
field
toward
−
x
,
so
the
minus
sign.
E₂
=
−8.99 ×
10
9
×
q₂
×
10
-6
÷
(
x₂
−
x
)
2
E₂
=
−8.99 ×
10
9
×
10
-6
÷
(0.3 − 0.1)
2
E₂
=
(−8.99
×
10
9
)
×
10
-6
0.2
2
E₂
=
(−8.99
×
10
9
)
×
10
-6
0.04
E₂
=
−
8,990
0.04
E₂
=
−224,750
N/C
Step 3 · Find net field at the point (E)
Use E = E₁ + E₂
Superposition: the two fields add as arrows along the line, signs and all.
E
=
3,596,000
+
(−224,750)
E
=
+3,371,250
N/C
Check
Put the numbers back in. Both sides should match.
+3,596,000
=
8.99 ×
10
9
×
4 ×
10
-6
×
0.1
÷
|0.1|³
✓
−224,750
=
8.99 ×
10
9
×
10
-6
×
(0.1
−
0.3)
÷
|0.1
−
0.3|³
✓
+3,371,250 = 3,596,000 + (−224,750) ✓
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.”
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.”