SME
Grade 11 Science · Problem type
The field of a point charge
Use this for “What is the electric field 0.30 m from a +2 μC charge, and the force on a 1 μC test charge there?”
Diagram
E
=
k|q|/
r
2
E
=
8.99 ×
10
9
×
2 ×
10
-6
/
0.3
2
E = 200,000 N/C
The field points away from a + charge.
Twice as far, a quarter of the field.
Test charge: 1 μC
Force on the test charge: 0.2 N
This is an example. Type your own values to start a new problem.
E (N/C)
=
8.99
×
10⁹
×
q (μC)
(
r (m)
)
2
q₀
Test charge
example · μC
F
Force on the test charge
calculated
Formulas
E
=
8.99 ×
10
9
×
q
×
10
-6
/(
r
2
)
2.00 ×
10
5
=
8.99 ×
10
9
×
2 ×
10
-6
/(
0.3
2
)
F
=
q₀
×
10
-6
×
E
0.2
=
1 ×
10
-6
×
(2.00
×
10
5
)
Assumptions
•
The field is the force on each coulomb of a small test charge: E = kq/r².
•
It points away from a + charge and toward a − charge (E < 0 here).
•
A test charge q₀ there feels F = q₀E; a negative q₀ is pushed against the field.
Step-by-step
WE KNOW
q = 2 μC, r = 0.3 m, q₀ = 1 μC
FIND
field (E), force on the test charge (F)
Step 1 · Find field (E)
Use E = 8.99 × 10⁹ × q × 10⁻⁶/(r²)
k
=
8.99 ×
10
9
N·
m
2
/
C
2
times
the
charge
in
coulombs,
over
r
2
.
E
=
8.99 ×
10
9
×
2 ×
10
-6
/(
0.3
2
)
q
=
2
μC
=
2 ×
10
-6
C
r
2
=
(0.3)
2
=
0.09
m²
E
=
8.99 ×
10
9
×
(2
×
10
-6
)/0.09
E
=
2.00 ×
10
5
N/C
Step 2 · Find force on the test charge (F)
Use F = q₀ × 10⁻⁶ × E
The field is newtons per coulomb: times the test charge in coulombs.
F
=
10
-6
×
(2.00
×
10
5
)
F
=
0.2
N
Check
Put the numbers back in. Both sides should match.
2.00 ×
10
5
≈
8.99 ×
10
9
×
2 ×
10
-6
/(
0.3
2
)
✓
0.2
=
10
-6
×
(2.00
×
10
5
)
✓
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