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utiaa a': "7A, Jalan utiaa a' 1/# (Tel: 16!712"$5/ 16!775$61#%Name : Teacher : Ms. Liow

Subject : Physics Second Term Class : U6 Physics

Chapter : ! Capacitance "# Lesson no: !

$ate: %%6

Time: .!&am '.!&pm

Topic (: Capacitance

(. Uses

. )s components in electronics and telecommunications.

(. *n radio and Tele+ision recei+ers and in transmitter circuits.

!. *n power supplies to smooth the recti,ication o, a.c. into d.c..

(.( Capacitors

. ) capacitor is a de+ice ,or storin- ener-y and electrical char-es. ut resultant char-e stored in a

parallel plate capacitor / &.

(. ) parallel plate capacitor consists o, two parallel metal plates separated by a thin dielectric

between the plates.

!. ) dielectric is an insulatin- material0 such as air0 oil0 mica0 paper or polystyrene.

$ielectric is used to increase the capacitance o, a capacitor.

1. Symbol :

(.! Capacitance0 C

. The capacitance0 C0 o, a body is de,ined as the ratio o, char-e stored to the potential o, the

body.

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QC

V=

0 2 / char-e stored on the body0 3 / potential o, body

(. The capacitance o, a parallel plate capacitor is:

QC

V=

where 2 / char-e stored on one plate0 3 / p.d between the plate.!. Unit o, capacitance is 4arad0 4

*n practice0 more common units are:6

&F F = 0(

&pF F= .1. $e,inition: ,arad "4# is the capacitance o, a parallel plate capacitor i, the char-e on either

plate is C when the potential di,,erence between the plates is 3.

5ample

"a# 7rite down an epression ,or the potential o, a metal sphere carryin- char-e 2.

"b# 8ence show that the capacitance C o, an isolated conductin- sphere is proportional to its radius

9.

"c# Calculate the capacitance o, the sphere i, its radius is (& m.

(.1 Parallel'plate Capacitor

. 5lectric ,ield stren-th inside the plates0

!&1;6 (

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VE

d=

where 3 / potential di,,erence between the plates

d / separation between plates

(. 4rom

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Problems

. 7hat is the capacitance o, two s>uare parallel plates 1.( cm on a side that are separated by

.? mm o, para,,in@

(. ) !&&p4 air-ap capacitor is connected to a !(3 battery. *, a piece o, mica ,ills the space

between the plates0 how much char-e will ,low ,rom the battery@

(.6 Capacitors in Series.

. Char-e on each capacitor is the same / 2

(. Since 2 / C30 the p.d. across each capacitor is:

QV

C=

(

(

QV

C=

!

!

QV

C=

!. Total potential di,,erence across the capacitors is:

( !

( !

Q Q QV V V V

C C C= + + = + +

( !

V

Q C C C = + +

The e>ui+alent or e,,ecti+e capacitance0 CT0 o, capacitors in series is:

T

V

C Q=

( !

TC C C C

= + +

!&1;6 1

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(.A Capacitors in Parallel

. Potential di,,erence across each capacitors is the same / 3

(. 4rom Q CV= 0 char-e on each capacitor are Q C V= ( (Q C V= ! !Q C V= .

!. Total char-e 2 supplied by the battery is:

( !Q Q Q Q= + +

( )

( !

( !

C V C V C V

C C C V

= + +

= + +

ut TQ C V=

The e>ui+alent or e,,ecti+e capacitance o, capacitors in parallel is:

( !TC C C C = + +5ample

4ind the combined capacitance o, each o, the ,ollowin- networBs0 assumin- each capacitor is ( F

!&1;6

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5ample (

4our capacitors each o, capacitance C are connected in such a way that their total capacitance is

also C. 7hich o, the ,ollowin- networBs -i+e this +alue@

5ample !

)n ?& F and a (& F capacitor are connected in series to a & 3 supply.

4ind: "a# the char-e on each capacitor "b# the p.d. across each capacitor.

!&1;6 6

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5ample 1

4ind the char-es 20 2(0 2!on the capacitors and the potential di,,erences 3and 3(across them.

(.6 5ner-y Stored *n a Char-ed Capacitor

. )t any instant durin- char-in-0 the char-e 2 on the capacitor increases as the potentialdi,,erence 3 o, the capacitor increase.

(. )s V Q 0 a -raph o, p.d. 3 a-ainst 2 is a strai-ht line passin- throu-h "&0.

!. 7orB done in char-in- the capacitor by dQ V dQ= / shaded area under -raph

1. The total worB done in increasin- the char-e ,rom Dero to 2 on the plates is:

0Q

o

QW V dQ V

C= =

. 8ence0 the ener-y stored in the capacitor

(

(

( ( (

QCV QV

C= = =

!&1;6 A

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5ample

) !& F capacitor is char-ed to !& 3 and a :& F capacitor is char-ed to 6 3. The batteries areremo+ed and the two capacitors are then connected with their plates o, liBe char-es to-ether.

"a# Calculate the new p.d. across each capacitor.

"b# Calculate the new char-e on each capacitor.

"c# 7hat is the loss o, electrical ener-y durin- this process@

7hat becomes o, this lost in potential ener-y@

(.A Char-in- ) Capacitor Throu-h a 9esistor "9'C char-in-#

. 7hen the contact o, the two'way switch E is at )0 the capacitor C will be char-ed by the

battery 3othrou-h the resistor 9.

(. 5lectrons are mo+ed ,rom plate a to plate b by the battery. Plate a becomes positi+ely char-ed0

with char-e F2 while plate b becomes ne-ati+ely with '2. ) current * ,lows throu-h the

capacitor.

!. The e>uations ,or current *0 char-e 2 and p.d. 3 durin- char-in- are:

!&1;6 ?

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a# Char-in- current0

t

RCoI I e

=

b# Char-e on capacitor0

t

RCoQ Q e

=

c# P.d. across capacitor0

t

RCo

V V e

=

7here *o/ initial char-in- current0 9C / time constant0

2o/ maimum char-e0 3o/ supply +olta-e

(.? $ischar-in- a Capacitor throu-h a 9esistor

. 7hen the contact E is switched to 0 electrons are trans,erred ,rom plate b to plate a. )

dischar-e current * ,lows.

(. The e>uations ,or *0 2 and 3 durin- dischar-e are:

a# dischar-in- current is:

t

RCoI I e

=

b# char-e on the capacitor0

t

RCoQ Q e

=

c# p.d. across capacitor0

t

RCoV V e=

!. 4rom

t

RCoV V e

=

ln ln ot

V VRC

= +

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(.; Time'constant0 9C

. Time'constant0 9C0 is the time taBen durin- char-in- ,or the char-e on a capacitor to increase0

,rom Dero to 6!G o, its maimum char-e.

4rom:

t

RCoQ Q e

=

7hen t / time'constant / 9C

RC

RC

oQ Q e

=

( )

( )

&.!A &.6! 06!G

o o

o o o

Q e Qe

Q Q Q

= =

= =

5-: 4ind the time'constant in the ,ollowin- circuits@

a.

b.

!&1;6 &

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Problems

. The capacitance in the circuit is C/ &.!& p4 0the total resistance is (& Band the battery

em, is (3 $etermine "a# the time constant "b# the maimum char-e the capacitor could

ac>uire "c# the time it taBes ,or the char-e to reach ;;G o, this +alue "d# the current Iwhen

the char-e Qis hal, its maimum +alue "e# the maimum current and ",# the char-e Qwhenthe currentIis &.(& its maimum +alue.

(. *n the 9C circuit shown in ,i-ure0 the battery has ,ully char-ed the capacitor0 soQ

0=C E .Then at t/ & the switch is thrown ,rom position a to b .The battery em, is (&.&

30 and the capacitance C / .&( 4. The currentIis obser+ed to decrease to &.& o, its

initial +alue in 1& s. "a# 7hat is the +alue o, 20 the char-e on the capacitor0 at t = &@ "b#

7hat is the +alue o, 9@ "c# 7hat is 2 at t/ 6& s@

!&1;6