Capacitance: Understanding Charge Storage and Energy in Capacitors, Study notes of Physics

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25-2. Capacitance

Capacitor:^

Any^ two^ isolated

conductors charge^ storage,

energy^ storage CV q^ =

Given As a Result

25-2. Capacitance 1 σ⎞⎛≡=⎟⎜^ εε ⎠⎝

q += EEE −+ A

ε ε⎠^00

A ⎝^ EdVV −=− if f

VVV

P^ i l (^

V l^ )^

VVif

V^ −=

Potential^ (or

Voltage)

q^ d A

EdV ==^ ε^0

A ε 0 C

Capacitance

0 C = d

C^ ⎤⎡

U it^ f^

it^

C ⎤⎡ ≡^ F ⎥⎢ V ⎦⎣

Units of capacitance: Farad^ (F) = Coulomb/Volt

25-2. Capacitance

CapacitanceCapacitanceand Your iPodand Your iPod

AQ and Your iPodand Your iPod^ ε^0^ C^ d

AQ V ε^0 C ==

25-2. Capacitance

-^ A huge parallel plate capacitor consistsg^

p^ p^

p of two square metal plates of side 50cm, separated by an air gap of 1 mmWh t i^ th

it^

-^ What is the capacitance?^ C^ =^ ε A/dC^ ε A/d^0 0^ = (8.85 x 10

–12^ F/m)(0.25 m

2 )/(0.001 m)

= 2 21 x 10

–9^ F

= 2.21 x 10

F (Very Small!!)

C^ ⎤⎡

U it^ f^

it^

C ⎤⎡ ≡^ F ⎥⎢ V ⎦⎣

Units of capacitance: Farad^ (F) = Coulomb/Volt

25-2. Capacitance

-^ A^ parallel

plate^ capacitor

of capacitance

C^ is^ charged

using^ a battery. • Charge = Q potential difference = V

–Q +Q

-^ Charge^ =

Q,^ potential

difference^

=^ V.

-^ Plate^ separation

is^ INCREASED

while battery^ remains

connected. Does the Electric Field Inside:(a) Increase?(b) R^ i^ th

S^?^ • V is fixed by battery!• C decreases (=

εA/d)^0

• Q=CV; Q decreases (b) Remain the Same?(c) Decrease?
• E = Q/^ εA decreases^0

25-3. Calculating the Capacitance Cylindrical CapacitorCylindrical^ Capacitor h 2 πε 0 = C = Cb ⎞⎛ ln^ ⎟⎜ a ⎠⎝

q ) 2 (π hrE = ε

== ∫∫

bq

qdr

drEV

b b

ln

ε

== ∫∫

ah

drrh

drEV

a a^

ln 2

0 0

πε πε

25-3. Calculating the Capacitance Spherical CapacitorSpherical Capacitor a = (^) C 4 πε C 4 πε 00 a^ ⎞⎛⎞⎛−^1 ⎟⎜ b ⎠⎝^ C^

4

ab >> ∞→ b

a C^ 4 πε=^0 Capacitance^ of^ a^ single

conducting

sphere

25-4. C in Parallel and in Series Parallel^

Series^

Q:^ the^ same

ParallelV:^ the^ same

25-4. C in Parallel and in Series^ Q: the same

Series^

Q^1

Q^2

Q:^ the^ same Series^ Q^ Q^

Q (WHY??)

A^ QQ^1 2 B^ C^ C

-^ Q=^ Q=^1

Q^ (WHY??)

•^ V=^ VAC^ AB

+^ V^ BC

CC^1

(^2) Q

QQ CC^21

Q^ += Ceq

C^ eq (^21)

eq^111 +=^ CCC

CC^21

Ceq

25-4. Example 1

10 F

What is the charge on each capacitor?

10 μF

-^ Q^ = CV; V =

120 V What^ is^ the

charge^ on^

each^ capacitor?

20 μF

Q^ C^ ;^

0

-^ Q=^ (10^ μF)(120V) = 1200^1

μC

-^ Q=^ (20^ μF)(120V) = 2400^2

μC Q^ (30^ F)(120V)

3600 C^

30 μF

-^ Q=^ (30^ μF)(120V)^3

= 3600^ μC

120V

Note that:•^ Total charge (

μC) is shared between the 3 capacitors in the ratio

C^ :C^ :C^ — i.e. 1:2:3^123

25-4. Example 3

What is the charge on the 10

μF capacitor? What^ is^ the

charge^ on^

the^10 μF^ capacitor?

10 μF 5 μF 5 μF^

10V 10 μF

-^ Then, we have two

^10 μ F^ capacitors in series

10 μF

-^ So, there is 5V across the

^10 μ F capacitor of interest

10 μF^

10V

-^ Hence,^ Q =

(^10 μ F^ )(5V) =

^50 μ C

25-4. Example 4

+ (N 1) Capacitors with A and d: Parallel Connection

(N-1) Capacitors with A and d: Parallel Connection

25-5. Energy Stored

1 2 CV

U^ =^2

CV

U

-^ Start out with uncharged capacitor•^ Start^ out

with^ uncharged

capacitor

-^ Transfer

small^ amount

of^ charge^ dq

from

one^ plate^ to

the^ other^ until

charge^ on h^ l^ h^

i^ d^ Q each^ plate^

has^ magnitude

Q

-^ How^ much

work^ was^ needed?

dq

Q = VdqU^ ∫

Q^ ==∫^

Qdq

q^

2

2 CV =

q ∫ 0

∫^

q C

C 0