SME
Grade 11 Science · Problem type
The field between two charges
Use this for “A +4 μC and a −2 μC charge are 0.5 m apart. What is the electric field 0.2 m from the +4 μC charge?”
Diagram
Field lines leave + charges and end on − charges; where they crowd, the field is strong.
At
x
=
0.2
m:
E₁
=
kq₁/
x
2
=
899,000
N/C
and
E₂
=
kq₂/
(x − r)
2
=
200,000
N/C,
each
+
pointing
toward
q₂’s
side.
E = E₁ + E₂ = 899,000 + 200,000
E = 1,100,000 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
First charge
example · μC
q₂
Second charge
example · μC
r
Distance between the charges
example
x
Distance of the point from q₁
example
E₁
Field from q₁ (+ toward q₂)
calculated · N/C
E₂
Field from q₂ (+ toward q₂)
calculated · N/C
E
Field at the point (+ toward q₂)
calculated · N/C
Formulas
E₁
=
8.99 ×
10
9
×
q₁
×
10
-6
/(
x
2
)
8.99 ×
10
5
=
8.99 ×
10
9
×
4 ×
10
-6
/(
0.2
2
)
E₂
=
−8.99 ×
10
9
×
q₂
×
10
-6
/(
(
r
−
x
)
2
)
2.00 ×
10
5
=
−8.99 ×
10
9
×
(−2)
×
10
-6
/(
(0.5 − 0.2)
2
)
E
=
E₁
+
E₂
1.10 ×
10
6
=
8.99 ×
10
5
+
(2.00
×
10
5
)
Assumptions
•
The point is on the line between the charges, x from q₁; + means toward q₂.
•
Each charge’s field points away from it if it is + and toward it if it is −.
•
The fields add: between unlike charges both point the same way; between like charges they partly cancel.
•
k = 8.99 × 10⁹ N·m²/C², and 1 μC = 10⁻⁶ C.
Step-by-step
WE KNOW
q₁ = 4 μC, q₂ = −2 μC, r = 0.5 m, x = 0.2 m
FIND
field from q₁ (+ toward q₂) (E₁), field from q₂ (+ toward q₂) (E₂), field at the point (+ toward q₂) (E)
Step 1 · Find field from q₁ (+ toward q₂): E₁
Use E₁ = 8.99 × 10⁹ × q₁ × 10⁻⁶/(x²)
q₁
’s
field
points
away
from
a
+
q₁
,
which
is
toward
q₂
,
so
it
is
+
when
q₁
is
+.
E₁
=
8.99 ×
10
9
×
4 ×
10
-6
/(
0.2
2
)
E₁
=
(8.99
×
10
9
)
×
(4
×
10
-6
)/0.04
E₁
=
35,960/0.04
E₁
=
8.99 ×
10
5
N/C
Step 2 · Find field from q₂ (+ toward q₂): E₂
Use E₂ = −8.99 × 10⁹ × q₂ × 10⁻⁶/((r − x)²)
The
point
is
r
−
x
from
q₂
.
A
+
q₂
pushes
back
toward
q₁
,
so
its
field
gets
a
minus
sign.
E₂
=
−8.99 ×
10
9
×
(−2)
×
10
-6
/(
(0.5 − 0.2)
2
)
E₂
=
(−8.99
×
10
9
)
×
(−2)
×
10
-6
/
0.3
2
E₂
=
(−8.99
×
10
9
)
×
(−2)
×
10
-6
/0.09
E₂
=
(1.798
×
10
10
)
×
10
-6
/0.09
E₂
=
17,980/0.09
E₂
=
2.00 ×
10
5
N/C
Step 3 · Find field at the point (+ toward q₂): E
Use E = E₁ + E₂
The two fields add as arrows along the line.
E
=
8.99 ×
10
5
+
(2.00
×
10
5
)
E
=
1.10 ×
10
6
N/C
Check
Put the numbers back in. Both sides should match.
8.99 ×
10
5
=
8.99 ×
10
9
×
4 ×
10
-6
/(
0.2
2
)
✓
2.00 ×
10
5
≈
−8.99 ×
10
9
×
(−2)
×
10
-6
/(
(0.5 − 0.2)
2
)
✓
1.10 ×
10
6
≈
8.99 ×
10
5
+
(2.00
×
10
5
)
✓
Topic check
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