Capacitance
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z
z
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Capacitor – device used to store
potential energy from an E field
The E field comes from stored
charge
This energy might be stored slowly,
but can be released quickly –
photoflash, heart defibrillator
A capacitor is formed from two
isolated conductors - equipotentials
When capacitor is charged, plates
have equal but opposite charges +q
and -q
Stroboscope
Capacitance
z
Capacitance is a proportionality
constant relating q and V
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q is the absolute value of the
charge on one plate.
V is the potential difference
between plates.
q = CV
z
or
C = q /V
C depends only on geometry of
plates, not on their q or V
Capacitance
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Capacitance is a measure of
how much q is needed on
plates to get V between them
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Greater C, more q required
SI unit for C is Farad
1F = 1C / V
Capacitance
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Can charge a capacitor using
a battery
Battery – device maintains
certain V between its
terminals by internal
electrochemical reactions
Initially V on plates is 0
Close switch, plates gradually
charge up to V of battery
through flow of electrons
Capacitance of Parallel Plates
We ignore these (edge) fringe fields
Capacitance (Exercise)
z
Does the C of a capacitor increase,
decrease or remain the same when
charge, q, on it is doubled
z B) V across it is tripled
z A)
Remember C of capacitor only depends on
its geometry so C is the same for A and B
Capacitance – Method of Calculation
z
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Calculate C of a capacitor from its geometry
using steps:
1) Assume charge, q, on the plates
r
r
2) Find E between plates
ε
E
•
d
A
=
0
using q and Gauss’ law
∫
3) Find V from E using
r
r
ΔV = − ∫ E • ds
f
i
z
4) Get C using
q
C =
V
qenc
Capacitance of Parallel Plates
z
Simplify Gauss’ law
z
1) Pick Gaussian surface to
enclose charge on + plate
and E and dA to be parallel
r
r
ε 0 ∫ E • d A = q enc
r
r
E • d A = EA
q = ε 0 EA
Capacitance of Parallel Plates
r r
ΔV = V f − Vi = − ∫ E • ds
f
i
z
2) For V choose path that
follows the E field line from
– plate to + plate then E and
ds are in opposite directions
r
r
E • d s = − Eds
+
V = V f − Vi =
∫ Eds
−
Capacitance of Parallel Plates
z
Find C for parallel plate
capacitor separated by d
z
E is constant between plates
+
V =
∫ Eds
−
z
d
= E
∫ ds
= Ed
0
A is area of plates
q = ε 0 EA
q ε 0 EA
C= =
V
Ed
Capacitance of Parallel Plates
z
Parallel-plate capacitor
C=
z
z
ε0 A
d
Only depends on area A of plates and
separation d
C increases if we increase A or decrease d
Hollow Conductor
z
Cylindrical capacitor
L
C = 2πε 0
ln(b / a )
z
Spherical capacitor
ab
C = 4πε 0
b−a
z
Isolated Sphere
C = 4πε 0 r
Capacitance (Exercise)
z
For capacitors charged by same battery, does q
stored by capacitor increase, decrease or remain
same if plate separation of parallel-plate
capacitor is increased.
q = CV
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All capacitors have same potential V from battery
and so q increases (decreases) with C
Capacitance
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If plate separation (d) of parallel
plate capacitor is increased,
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d increases so C decreases
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C decreases so q decreases
q = CV
C=
ε0 A
d