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33 terms

physics test 2

STUDY
PLAY
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)
746 w
1hP
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