James Hyatt – Physics 1E03 Crib Sheet

James Hyatt – Physics 1E03 Crib Sheet Outline I. Electrostatics (4 weeks) II. DC Circuits (2 weeks, plus labs) III. Magnetism (3 weeks) IV. Waves (3 weeks) -includes electromagnetic waves

****************************************** I.

Electrostatics (4 weeks)

The “source” q produces an electric field that pushes test charge q0 Units=N/C *E=F/q0

 q E = k e 2 rˆ r

**The field decrases in strength with distance **Fields add together with multiple charges. Vector Sum

Chapter 23 e ≈ +1.602×10-19 C Insulator: can be charged by rubbing but charges DO NOT move (ie. glass, rubber, paper) Conductors: (some) charges move freely (ie. metals, some liquids) Semiconductors: electrical properties are between insulators and conductors (ie. Si, Ge)

2 r qq F =k rˆ ke = 8.988 x109 N • m 2 C r 1

12

e

2

2

*K = “Coulomb’s Constant” **where r hat is a unit vector parallel to r** When one q is doubled and one is halved, the angle does not change. If the masses were to change, the angle would too.

Lambda is linear charge density (charge per length of the object). L is S for a curve **Integrate between d and d+L. Set q equal to lambda dx, you know that lambda and k are constant so they get removed from the integral. Integrate 1/x2, then plug in d and d+L. (Keep in mind b-a for integrals)**

Given m=1g, L=60cm,q1=q2 Solution: FBD, Fg=Ftsinx, find Ft, get Ftcosx, trig=find distance between q, Fqq=Ftcosx=(kqq)/r^2

In this example, be sure to consider the force between q3 and q4 (ie, watch the diagonal force as well). Electric Fields

**use pythag to note that r2 = a2 + x2, multply the integral of dE by costheta (since it is a constinuous charge distribution). The integral of dq is Q because of s=r*theta and dq=lambda*ds and the integral between 0 and 2Pi of lambdaRdtheta equals lambda 2Pir which equals Q because 2Pi = theta and r = r**

**integrate dE=dEcostheta to get to klambda/r as a constant is multiplied by the integral of costhetadtheta**

Gauss

**Note that there are twice as many field lines on the 2Q as opposed to the Q because # of field lines are proportional to charge**

**Flux is equal to E dot A. A is 4Pi*r2-. This is all equal to q(enclosed) over epsilon. Rearrange for E. Note that 1/4Pi*epsilon is equal to k. This gives you E = k*q/r2**

Uniformly charged sphere: Since one can decide the Gaussian surface themselves, let’s analyze what would happen if the radius of the surface is less than the original and greater than it.

**E is parallel to dA and is constant on the Gaussian surface S, therefore can take it out of the integral as a constant. The integral of dA is just A. The area of a sphere is 4Pi*r2. Rearrange for E, get E=k*q/r2 NOTE: Gaussian radius is little r, actually radius is big R** **For when the Gaussian surface is smaller than the surface of the sphere the equation ends up being E=kqr/R3. This was derived using the volume charge density. Since the exact Qenclosed is not known, it is used to solve for this unknown variable. The E is still parallel and constant.** For any uniform distribution of charge Q within a sphere of radius R:

Conductors in Electrostatic Equilibrium (F=0!): -There is NO electric field INSIDE of a conductor -The field lines are perpendicular to the surface -Any and all net charge resides on the surface ONLY

V=J/C Work

**For P1 there is no flux because it is inside of the conductor and there is no charge that resides within a conductor. For P2 it works out to just E=k*q/r2** *Any Gaussian surface inside the conductor encloses zero net charge.

W=Fd=qEd,

W=K=1/2*m*v^2

**Electric potential is absolute, no –ve, EVER!** Electron Volt Equipotential: a surface on which V is constant. They are always perpendicular to the electric field, a conductor is always an equipotential, the closer they are, the larger the E, the farther they are, the weaker the E.

**Derive V with respect to x!! DO NOT try and derive x. Recall with dA you just take A, same idea here

**V=Ed as well as dV=Eds, the +ve plates have the high potential of 20V whereas theve plates have the low potential of 0V**

Electric Potential (V)

Solution to above picture **Note, with squares or multiple sided things, ensure to ADD THE DIAGONALS!!**

*Potentials add as SCALARS!

**Inside of a conductor the electric field is 0 however the electric potential is a constant. As you approach the edge and leave it it decreases. Note that the electric field increases and the V decreases as it approaches R**

END OF FIRST MIDTERM MATERIAL Capacitance

Finding C for symmetrical charges 1) Find the electric field as a function of r using Gauss’s Law. 2) Choose a path along a field line: the potential change with each small change dr in distance is dV = - E(r)dr 3) Integrate from start to end of field line to find the potential difference. 4) C=Q/V

**C depends only on the geometry of the plates

**Lambda=Q/L Integrate V= - int(Eds) using lambda for Q, between r1 and r1 Sub Q/L for lambda, plug into C=Q/V Simplify, sub ebsalom-not for k-not Dielectric Materials

 Plates moved apart: -field same -potential increase (V=Ed) -capacitance decrease

 Derive: 1. E=kQ/r^2 2. V=-INTEGRAL(E*ds) =-INT:R1-R2([kQ/r^2]*ds) = kQ[1/R1 – 1/R2] 3. C=Q/V =(1/k)[1/R1 – 1/R2]^-1 (For parallel plates) “Kappa”=Dielectric Constant C=(1/k)[1/R1 – 1/R2]^-1 ^Finding C between two spheres in one another, like above, and Earth/Ionosphere

Coaxial: E=(2K”lambda”)/r V=2k(“lambda”)*ln(b/a) Max poential diff of inner/outer part of cable with dielectric: V=Emax*a*ln(b/a) **(B is outer radius, A is inner radius)

**C=Q/V=Q/(Ed)

II. Circuits

**Solution Q1=CV=120mC (120-Q2)/6=Q2/3 Q2=40mC Q1=80mC **Energy Stored in a Capacitor

U=(1/2)E0AE^2d (For parallel plates capacitor) ****No longer in ELECTROSTATICS, now having current where E0 is NOT zero inside of a conductor!!!!!!!!!****

Parallel

Series

Current Density (J) Defined as current per unit area of a conductor

Electrical Work and Power Electrical resistance converts electrical potential energy to thermal energy (heat),

(“sigma” is the conductiviy, not charge density here) **J and Eo are created as a result of peotential differences across the conductor.

OR

In a condcuting wire: **Ya! :D

SO BASICALLY THIS “E” SYMBOL IS VOLTS!!!!!!

Summary

#YOLO (Jokes ;) Brightening up your studying!!!! :D ) DC Circuits

DC=Current flowing in one direction AC=Alternating current

Solution: Change in V in loop = 0 #1: 12V-0.01*I1-1*I2-10V=0 =2V-0.01I1-I2 #2: 10v+1I2-0.96I3=0 I3=I1-I2 6*#1+#2= I1=171.95A I2=0.283A I3=171.67A RC Circuits “Circuits in which currents vary in time”

III. Magnetism

Question:

Solution V(t)=V0e^-t/T 12V=18Ve^-10/T Ln(2/3)=-10/T T=24.7seconds=RC

Charging :

Q=charge, v=velocity, B=Magnetic Field(T)

******NOTE THIS!!!!!!******

NOTE

Magnetic Forces and Torques

Torque on a current loop

****LORENTZ FORCE!!!!!******** (Use the force Luke! The force is strong with this one ;D )

**Note below***

W= (use to get frequency with f=w/2Pi)

KE=(q^2*B^2*r^2)/(2*m) **Since v=(qBr)/m Velocity selector “Field of a current element” (Instantanious at a point in space)

***Int(B*ds)=u0int(J*dA)  ** JdA=I ** Magnetic Flux

Units= 1 Weber (Wb)

This is like Gauss’ law, but for magnetism. However, it states that: - the number of magnetic field lines that enter a volume enclosed by a surface S must equal the number that leave the volume -and it implies that magnetic monopoles do not exist Question:

Solution: JA=I Iu=int(Bds) Iuds=B Solenoid:

Faraday’s Law: When the external magnetic flux ΦB through a closed conducting loop with one turn changes, the emf induced in the closed loop is:

Self Inductance

***UP TO LECTURE 25***

Solution: Emf(AKA V)=-L(dI/dt)=0 dI/dt=V/L int-0tot(dI)=int-0tot([V/L]dt) I(t)=E*t/L I=12V*2ms/20mH

Total Potential Energy stored by an Inductor :