Essay on 204 Physics Exam1Cheatsheet 1 1

Submitted By am1418
Words: 1374
Pages: 6

Proton mass = 1.673e−23kg
Neutron = 1.675e−27 kg
Electron = 9.11e−31kg

Magnitude of a charge:

v 2 = vo2 + 2ax v = vo = at vo = v 2f − 2ax

e = 1.6e−19 C

1C = 6.25e C q = Ne, N=#electrons

Parallel Plate Capacitor

18

Charge:

Law of conservation of Electric Charge
The net electric charge of an isolated system remains constant • Rub wool on plastic rod – electrons transferred onto plastic rod
• Rub wool on glass – electrons on glass is stripped away • Uncharged wood and charged glass are attracted
(polarization effect) Coulomb’s Law:

 q q
F=k 1 22 r , 1µC

= 10 −6 C

Nm 2
1
k = 8.99e
=
C2
4πε 0

stored A

Permitivity of Free Space:

1
C2
= 8.85e−12
4π k
Nm 2

stored: B
• Gravitational Force = always attractive

Fg =

Gm1 m2
Nm 2
, G = 6.7e−11 2
2
r kg q
, σ =
A

same as if ALL the charge q were concentrated at a point charge at the CENTER of the shell
• The only charge in the Gaussian surface is the charge q on the shell ∴ Q=q and:

(charge density)

q
• “E” has the same magnitude between the plates
(for r>R)
• E= except at the edges
2
4 πε or
• Field is NOT dependent on distance between the charges • “r”=distance from center to Gaussian surface Electric Field Lines
⊕q⇒lines radially OUTward
q ⇒ lines radially Inward
• Lines always from positive toward negative
• # of lines leaving a charge is ∝ to the magnitude of the charge a)


Mistakes:
1. Two field lines cross at P
2. # of lines are not correct: a total of 8 lines leave the
+4q, ∴ 4 of those lines must enter the ‐2q charge
3. Lines should not be evenly spaced (this isn’t in a parallel plate capacitor)
Electric Dipole: two separated charges of SAME magnitude, but OPPOSITE signs The E inside a Conductor: Shielding
• At equilibrium under electrostatic conditions, any
EXCESS charge resides on the SURFACE of the conductor • Interior is electrically neutral (NO E inside)

9

ε0 =

q σ E=
=
εo A εo




Fcent = Fel

(G is negligible compared to Fel)

rFc m 
Find Tension and E
Given: m, q, θ



 Fe
E=
N/C q0 →

Fe = qo E

Σ

Gaussian Surface
Uniform/Symmetrical Charge distributions and the
Electric Field







Electric field by a point charge q:

kq
E= 2 r ← Does not depend on q0

Q1
±d =
(r − d) q2 Gauss’ Law for a point charge:

Canceling of Electric Fields


 
E = 0 where E1 = E2

Q1 q2 =
, Q1 = larger charge d 2 (r − d)2

EA =

• The value of “d” that’s NOT in between the two charges is the WRONG value.

• The location where E = 0 will be closer to the smaller charge Symmetry and the Electric Field

q
= ElectricFluxΦ E εo • Where E is the electric field at ANY point on the surface • Direction of E is NOT necessarily ⊥ to the Gaussian
Surface
• Magnitude of E can vary from point to point on the surface Gauss’ Law for a Gaussian Surface

Φ E = Σ ( E cos φ ) ΔA =

Φ E = E(ΣΔA)

Q Nm2 εo C

(ΣΔA) = Area of Gaussian Surface

Φ E = E(4π r 2 ) = kq4π =

does NO WORK if it moves ON the surface:
AB

WAB EPE A EPE B
=
− qo qo qo V = EPE J/C qo A

o



surfaces (different radii)



EPE B EPE A ΔEPE −WAB

=
=

qo qo qo qo 
E is ⊥ to the Equipotential Surface (not necessarily

a sphere); it points in the direction of DECREASING
POTENTIAL (farther away from the charge)
• E ⊥ ⇒ Fe does No Work ∴ ΔV=0
• Conductors at EQUILIBRIUM under Electrostatic
Conditions
• Parallel Plate Capacitors: ΔV between plates:

WAB = qo EΔs

If ΔV is  ⇒ VA>VB
If ΔEPE is  ⇒ EPEA>EPEB
⊕q’s accelerate from higher V  lower V
q’s accelerate from lower V  higher V
Positive charges flow→ negative terminals

, V BETWEEN the plates:

V = −qo EΔs qo

and

E = − ΔV = potential gradient
Δs

(only the Component of E along Δs (NOT the ⊥ component ∴ does work)

CAPACITORS AND DIELECTRICS
EPE A − EPE B
Dielectric: an electrically insulating material between the qo = conductors
VA − VB
•…