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Electrical Physics 1E03 Summary Exam Sheet.doc

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Waldemar Okon

James Hyatt – Physics 1E03 The “source” q produces an electric field that pushes test charge q0 Crib Sheet Outline  q I. Electrostatics (4 weeks) Units=N/C E = k e 2rˆ *E=F/q 0 r II. DC Circuits (2 weeks, plus labs) III. Magnetism (3 weeks) IV. Waves (3 weeks) -includes electromagnetic waves **The field decrases in strength with distance ****************************************** **Fields add together with multiple charges. I. Electrostatics (4 weeks) 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) r qq 9 2 F 12k e 1 2rˆ ke=8.988x10 N• m 2 r 2 C Lambda is linear charge density (charge *K = “Coulomb’s Constant” per length of the object). L is S for a curve **where r hat is a unit vector parallel to r** When one q is doubled and **Integrate between d and d+L. Set q equal to one is halved, lambda dx, you know that lambda and k are the angle does constant so they get removed from the integral. not change. If Integrate 1/x , then plug in d and d+L. (Keep in the masses mind b-a for integrals)** were to change, the angle would too. 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 **use pythag to note that r = a + x , multply the force between q a3d q (ie4 watch the diagonal force as well). 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 Electric Fields 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** 2 **Flux is equal to E dot A. A is 4Pi*r -. This is all equal to q(enclosed) over epsilon. Rearrange for E. Note that 1/4Pi*epsilon is 2 equal to k. This gives you E = k*q/r ** 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*r .2 2 Rearrange for E, get E=k*q/r 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/R . 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 Conductors in Electrostatic Equilibrium within a sphere of radius R: (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 P1there is no flux because it is inside W=Fd=qEd, W=K=1/2*m*v^2 of the conductor and there is no charge that resides within a conductor. For P 2t works out to just E=k*q/r ** *Any Gaussian surface inside the conductor encloses zero net charge. **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 the- ve 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
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