SME
University Physics II: Electricity & Magnetism
Electric fields and Gauss's law
Use this for “A ball 0.2 m in radius holds 3 μC spread evenly through it. Find the field and the flux 0.1 m and 0.5 m from its center.”
Diagram
E
×
4πr²
=
Q(inside)
ε₀
,
so
E
=
kQ(inside)
r
2
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
2
=
200,000
N/C
Φ
=
Q(inside)
ε₀
=
226,000
N·
m
2
/C
This is an example. Type your own values to start a new problem.
E (N/C)
=
k
Qenc (
)
(
r (m)
)
2
Q
Charge on the ball
example
R
Radius of the ball
example
Φ
Flux through the sphere
calculated · N·m²/C
Formulas
Qenc
=
Q
2 ×
10
-6
=
2 ×
10
-6
E
=
8.99 ×
10
9
×
Qenc
r
2
1.9978 ×
10
5
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
2
Φ
=
4π
×
8.99 ×
10
9
×
Qenc
2.2594 ×
10
5
=
4π
×
8.99 ×
10
9
×
(2
×
10
-6
)
Formulas are worked in C. The step-by-step shows the conversions.
Assumptions
•
The charge is spread evenly through the ball, so E points straight out and is the same all over a sphere of radius r.
•
Gauss’s law: the flux through that sphere is Φ = E × 4πr² = Qenc ÷ ε₀, with 1 ÷ ε₀ = 4πk and k = 8.99 × 10⁹ N·m²/C².
•
Inside the ball the sphere holds the share (r ÷ R)³ of Q, so E = kQr ÷ R³; outside it holds all of Q, as a point charge would.
Step-by-step
WE KNOW
Q = 2 μC, R = 0.1 m, r = 0.3 m
FIND
charge inside r (Qenc), electric field at r (E), flux through the sphere (Φ)
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 = 2 μC = 2 × 10⁻⁶ C (1 C = 10⁶ μC)
Step 2 · Find charge inside r (Qenc)
Use Qenc = Q
r
is
outside
the
ball
(
r
≥
R
):
the
sphere
holds
all
of
the
charge.
0.3 ≥ 0.1: outside
Qenc
=
2 ×
10
-6
C
Step 3 · Find electric field at r (E)
Use E = 8.99 × 10⁹ × Qenc ÷ r²
E
×
4πr²
=
Qenc
ε₀
and
1
4πε₀
=
k;
a
negative
E
points
in.
E
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
2
E
=
(8.99
×
10
9
)
×
2 ×
10
-6
0.09
E
=
17,980
0.09
E
=
1.9978 ×
10
5
N/C
Step 4 · Find flux through the sphere (Φ)
Use Φ = 4π × 8.99 × 10⁹ × Qenc
Gauss’s
law:
the
flux
is
the
charge
inside
over
ε₀,
and
1
ε₀
=
4πk.
Φ
=
4π
×
8.99 ×
10
9
×
(2
×
10
-6
)
Φ
=
2.2594 ×
10
5
N·
m
2
/C
Step 5 · Convert the answers
Change each answer back to the unit you chose.
Qenc = 2 × 10⁻⁶ C = 2 μC (1 C = 10⁶ μC)
Check
Put the numbers back in. Both sides should match.
2 ×
10
-6
=
2 ×
10
-6
✓
1.9978 ×
10
5
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
2
✓
2.2594 ×
10
5
=
4π
×
8.99 ×
10
9
×
(2
×
10
-6
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
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.”