# physics test 2

### 33 terms by raisin16 Plus

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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.

PEb-PEa=-qEd
W=-qEd

=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

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

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

1.6 x 10^-19 J

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= Eo/E

cd/Eo

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

V=Ed

### 17.9 Energy density

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

### Electric current CH.18

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

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)]

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)

1hP

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

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

Example: