V Q Q v Q v v F r ˆ r ˆ - Northeastern University

Force and Field

v

F12

=

1 4 0

Q1Q2 r2

r^

v

E1

=

1 4 0

Q1 r2

r^

v

F12

v

= Q2E1

= Q2

1 4 0

Q1 r2

r^

Pot. Energy and Voltage

U12

=

1 4 0

Q1Q2 r

V1

=

1 4 0

Q1 r

U12

=

Q2V1

=

Q2

1 4 0

Q1 r

Potential energy: U = -WField = WExternal

Gauss's Law:

rr

EdA

=

Qinside

0

Field and Voltage:

r

E

=

-

V x

,-

V y

,- V z

In a uniform field: V = -Ed

Capacitance: C = Q V

Parallel plate capacitor:

C

=

0

A d

;

E

=

0

Capacitor in series: 1 =

1

Ceq i Ci

Capacitors in parallel: Ceq = Ci i

Energy: U = Q2 = 1 CV 2 = 1 QV and u = 1 E2

2C 2

2

2

Current:

I

=

dQ dt

=

n

|q

|

vd

A

and

j

=

I A

=

n | q | vd

Ohm's Law: V = IR and E = j = j

Power: P = IV = I 2R = V 2 R

with R = L A

Resistors in series: Req = Ri i

Resistors in parallel:

1 = 1

Req i Ri

Kirchhoff's rules: I = 0 (Junction rule) ; V = 0 (Loop rule)

Capacitor charging:

Q

=

Qf

1

-

-t

e RC

and

-t

I = I0e RC

with Q f

= VC;

I0

=

V R

Kirchhoff's conventions:

If your loop goes through a battery from ? to + the Voltage increases (e.g. V = +12 V)

- +

V=12V

If your loop goes through a battery from + to ? the Voltage decreases (e.g. V = -12 V)

- +

V=12V

2

If you go across a resistor and the loop direction and

guessed current direction are the same, the voltage

i

decreases (e.g. V = -iR)

2

If you go across a resistor and the loop direction and guessed current direction are opposite, the voltage increases (e.g. V = +iR)

i

Units and constants:

0 = 8.854 ?10-12 C2/N.m2

and 1 = 9.0 ?109 N.m2 / C2 4 0

e = 1.602 ?10-19C

[Charge]=Coulomb=C [Electric Force]=Newton=N [Electric Field]=[Force]/[Charge]=N/C=[Voltage]/[distance]=Volt/meter= V/m [Energy]=Newton.meter=Joule=J [Voltage]=[Energy]/[Charge]=J/C=Volt=V [Capacitance] = [Charge]/[Voltage]=Coulomb/Volt=C/V=Farad=F [Current] = [Charge]/[time]=Coulomb/sec.=C/s=Amp=A [Resistance] = [Voltage]/[Current]=Volt/Amp= Ohm=

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