SME
University Physics II: Electricity & Magnetism · Problem type
Potential near a charged sphere, and the work to move a charge
Use this for “A metal sphere holds +5 μC. How much work does it take to bring a 3 nC charge from 0.5 m to 0.2 m from its centre?”
Diagram
V₁
=
kQ
r₁
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
V₁ = 59,930 V
V₂
=
kQ
r₂
=
8.99 ×
10
9
×
2 ×
10
-6
0.1
V₂ = 179,800 V
ΔV = V₂ − V₁ = 179,800 − 59,930
ΔV = 119,900 V
W
=
qΔV
=
1 ×
10
-9
C
×
119,900
V
W = 0.0001199 J
Carrying q round a dashed circle takes no work: only the change from r₁ to r₂ counts.
This is an example. Type your own values to start a new problem.
W (J)
=
q (nC)
k
Q (μC)
r₂ (m)
−
k
Q (μC)
r₁ (m)
V₁
Potential at r₁
calculated · V
V₂
Potential at r₂
calculated · V
ΔV
Potential difference V₂ − V₁
calculated · V
Formulas
V₁
=
8.99 ×
10
9
×
Q
×
10
-6
r₁
+5.9933
×
10
4
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
V₂
=
8.99 ×
10
9
×
Q
×
10
-6
r₂
+1.798
×
10
5
=
8.99 ×
10
9
×
2 ×
10
-6
0.1
ΔV
=
V₂
−
V₁
+1.1987
×
10
5
=
1.798 ×
10
5
−
(5.9933
×
10
4
)
W
=
q
×
10
-9
×
ΔV
+1.1987
×
10
-4
=
1 ×
10
-9
×
(1.1987
×
10
5
)
Assumptions
•
Outside a uniformly charged sphere, V is that of a point charge Q at its centre: V = kQ ÷ r, with k = 8.99 × 10⁹ N·m²/C² and V = 0 far away.
•
Both points are outside the sphere, and r₁ and r₂ are measured from its centre.
•
Carrying q slowly from r₁ to r₂ takes work W = qΔV = q(V₂ − V₁), the change in its potential energy.
•
The test charge q is small, so it doesn’t move the charge on the sphere.
Step-by-step
WE KNOW
Q = +2 μC, r₁ = 0.3 m, r₂ = 0.1 m, q = +1 nC
FIND
potential at r₁ (V₁), potential at r₂ (V₂), potential difference V₂ − V₁ (ΔV), work to move q (W)
Step 1 · Find potential at r₁ (V₁)
Use V₁ = 8.99 × 10⁹ × Q × 10⁻⁶ ÷ r₁
k
times
the
charge
in
coulombs
(μC
×
10
-6
),
over
the
distance
r₁
from
the
centre
(not
r₁
2
).
V₁
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
V₁
=
17,980
0.3
V₁
=
+5.9933
×
10
4
V
Step 2 · Find potential at r₂ (V₂)
Use V₂ = 8.99 × 10⁹ × Q × 10⁻⁶ ÷ r₂
k
times
the
charge
in
coulombs
(μC
×
10
-6
),
over
the
distance
r₂
from
the
centre
(not
r₂
2
).
V₂
=
8.99 ×
10
9
×
2 ×
10
-6
0.1
V₂
=
17,980
0.1
V₂
=
+1.798
×
10
5
V
Step 3 · Find potential difference V₂ − V₁ (ΔV)
Use ΔV = V₂ − V₁
The
difference
runs
from
the
start
to
the
end:
the
potential
where
q
ends
less
where
it
starts.
ΔV
=
1.798 ×
10
5
−
(5.9933
×
10
4
)
ΔV
=
+1.1987
×
10
5
V
Step 4 · Find work to move q (W)
Use W = q × 10⁻⁹ × ΔV
A
volt
is
a
joule
per
coulomb,
so
the
charge
in
coulombs
(nC
×
10
-9
)
times
ΔV
gives
the
work
in
joules.
W
=
10
-9
×
(1.1987
×
10
5
)
W
=
+1.1987
×
10
-4
J
Check
Put the numbers back in. Both sides should match.
+5.9933
×
10
4
=
8.99 ×
10
9
×
2 ×
10
-6
0.3
✓
+1.798
×
10
5
=
8.99 ×
10
9
×
2 ×
10
-6
0.1
✓
+1.1987
×
10
5
=
1.798 ×
10
5
−
(5.9933
×
10
4
)
✓
+1.1987
×
10
-4
=
10
-9
×
(1.1987
×
10
5
)
✓
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