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conductors

are highly charged, charges flows freely, electrons are bound loosely.
eg. nail

insulator

almost no charge flows,fewer force elctrons eg. wood

semiconductor

fewer free electrons

Electroscope

detects chargy by induction and conduction, electroscope contains a conductors.

Electric potential ch.16

PEb-PEa=-qEd
W=-qEd

Electric potential Va ch.16

=Pea/q

Electric potential ch.16 Vba=Vb-Va

=PEb-PEa/q= -Wba/q
W= Fd
F=qE
v=w/q; V=Ed
E=delta V/d
E=-Vba/d

Coloumbs law ch.16

F= k Q1Q2/r^2
Fr= mv^2/r
kq^2/r= mv^2/r
mv^2r=kq^2
r=kq^2/mv^2

Electric field ch.16

E=F/q

number 59; ch.16

Fe=lqlE= mg=neE=4/3 pi r^2 pq
n=4 pi r^3 q/ 3eE

Point charge ch.16

E=Q/r^2= 1/4pi Eo
lfl=kQq/r^2
E=kQ/r^2= 1/ 4 pi Eo

1 eV

1.6 x 10^-19 J

Electron Volt ch.17

v=w/q
w=qv

Electron Volt

V=k Q/r; = 1/4 pi Eo-Q/r

Capacitance ch.17

C= Eo/d (A); C=KEo (A/d)
Q=CV
C=Q/V

Eo

8.85 x 10^-12 c^2/nm^2

K ch. 17

K= Eo/E

Area of capacitance=

cd/Eo

Energy ch.17

U=1/2 QV
Q=CV
U=1/2(CV)V
U=1/2CV^2
V=Q/C
U=1/2 Q^2/C

Energy ch. 17 E=V/d

V=Ed

Energy ch.17 volume

Ad

17.9 Energy density

u= U/volume
u= Uvol= 1/2 CV^2/VOL
= 1/2 EoA/d (v^2/vol)
=vol= Ad
1/2 Eo A/d (Ed/ Ad)
u=1/2 EoE^2

Electric current CH.18

I= delta Q/ delta t
I= c/s
1 c/s= 1A

Ohms law ch.18

I proportional v
V=IR
R=V/I

chapter 18 Reseitance and ro

pT=po [1 + alpha (T-To)]
R=p L/A
Rt=Ro[1 + alpha (T-To)]

Resistance ch.18

R=p L/A
RA= pL
p=RA/L

Electric power chapter 18

P=IV= I (IR)= I^2 R
P=IV=(V/R)V= V^2/R

Electric power chapter 18

P= energy transformed / time
=q/t x v
P=IV

Alternatinv Current-chapter 18

P=Irms Vrms= Ipeak/ square root of 2 Vrms
= square rt of 2 (P)/vRMS
: P=1/2 Io^2 R= l^2rmsR
:P=1/2 Vo^2/R= V^2rms/R
: Vo=square rt 2 (Vrms)

Alternating Current- Peak current ch.18

I=V/R

Alternating Currents Chapter 18; I^2

=1/2 Io^2 and V^2=1/2 Vo^2
Irms= square route of I-> Io/square root of 2

Vrms-> Square root of V-> Vo/Square root of 2

Capacitance of an axon area of cylinder; chapter 18

A= 2pi r L
c= KEo A/d
Q=cv
v=pi r^2 L

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