SME
University Physics II: Electricity & Magnetism · Problem type
A dielectric slid into a charged capacitor
Use this for “A 25 pF capacitor is charged to 60 V and disconnected, then a slab with κ = 3 is slid in. Find the new voltage and energy.”
Diagram
Q = C₀V₀ = 10 pF × 100 V = 1,000 pC, and it stays: the battery is gone
C = κC₀ = 4 × 10 = 40 pF
V
=
Q
C
=
V₀
κ
=
100
4
=
25
V
U
=
Q
2
2C
=
U₀
κ
=
5 ×
10
-8
J
4
U
=
1.25 ×
10
-8
J
Disconnected: Q can’t change, so the slab lowers V and U; the field pulls the slab in and does the difference as work.
This is an example. Type your own values to start a new problem.
V (V)
=
V₀ (V)
κ
C₀
Capacitance before the slab
example · pF
Q
Charge on the plates
calculated · pC
C
Capacitance with the slab
calculated · pF
U₀
Energy before the slab
calculated · J
U
Energy with the slab
calculated · J
Formulas
Q
=
C₀
×
V₀
1,000 = 10 × 100
C
=
κ
×
C₀
40 = 4 × 10
V
=
V₀
κ
25
=
100
4
U₀
=
1
2
×
C₀
×
10
-12
×
V₀
2
5 ×
10
-8
=
1
2
×
10 ×
10
-12
×
100
2
U
=
Q
2
2 ×
C
×
10
-12
1.25 ×
10
-8
=
1,
000
2
2 × 40
×
10
-12
Assumptions
•
The capacitor is charged to V₀ and then the battery is removed, so the charge Q = C₀V₀ has nowhere to go and stays the same.
•
The slab fills the whole gap, so C = κC₀; with Q fixed, V = Q ÷ C = V₀ ÷ κ.
•
The stored energy U = Q² ÷ (2C) falls by the factor κ: the field pulls the slab in and does the difference as work.
•
pF × V gives pC, and pC² ÷ pF gives pJ, so × 10⁻¹² gives joules.
Step-by-step
WE KNOW
C₀ = 10 pF, V₀ = 100 V, κ = 4
FIND
charge on the plates (Q), capacitance with the slab (C), voltage with the slab (V), energy before the slab (U₀), energy with the slab (U)
Step 1 · Find charge on the plates (Q)
Use Q = C₀ × V₀
The
charge
the
battery
left
on
the
plates
before
it
was
removed;
pF
×
V
gives
pC.
Q
=
10
×
100
Q
=
1,000
pC
Step 2 · Find capacitance with the slab (C)
Use C = κ × C₀
The
slab
weakens
the
field
between
the
plates
by
κ
,
so
the
same
charge
sits
at
κ
times
less
voltage:
C
grows
by
κ
.
C
=
4
×
10
C
=
40
pF
Step 3 · Find voltage with the slab (V)
Use V = V₀ ÷ κ
Q
stays
the
same
and
C
is
κ
times
bigger,
so
V
=
Q
C
is
κ
times
smaller.
V
=
100
4
V
=
25
V
Step 4 · Find energy before the slab (U₀)
Use U₀ = ½ × C₀ × 10⁻¹² × V₀²
The
energy
before
the
slab:
half
C₀
in
farads
(pF
×
10
-12
)
times
V₀
2
.
U₀
=
1
2
×
10 ×
10
-12
×
100
2
U₀
=
1
2
×
(10
×
10
-12
)
×
10,000
U₀
=
(5
×
10
-12
)
×
10,000
U₀
=
5 ×
10
-8
J
Step 5 · Find energy with the slab (U)
Use U = Q² ÷ (2 × C) × 10⁻¹²
With
the
charge
fixed,
use
the
form
with
Q
:
pC²
pF
gives
pJ,
and
×
10
-12
gives
joules.
U
=
1,
000
2
2 × 40
×
10
-12
U
=
1,
000
2
80
×
10
-12
U
=
1,000,000
80
×
10
-12
U
=
1.25 ×
10
-8
J
Check
Put the numbers back in. Both sides should match.
1,000 = 10 × 100 ✓
40 = 4 × 10 ✓
25
=
100
4
✓
5 ×
10
-8
=
1
2
×
10 ×
10
-12
×
100
2
✓
1.25 ×
10
-8
=
1,
000
2
2 × 40
×
10
-12
✓
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