Electrostatics - A review

1 P13-Class 13: OutlineHour 1:Concept Review /OverviewPRS Questions –possible exam questionsHour 2:Sample ExamEXAM Thursday: 7:30 –9 pm2 P13-Exam 1 Topics•Fields (visualizations)•Electric Field & Potential•Discrete Point Charges•Continuous Charge Distributions•Symmetric Distributions –Gauss’s Law•Conductors•Capacitance•Calculate for various geometries•Effects of dielectrics•Energy storage3 P13-General Exam Suggestions•You should be able to complete every problem•If you are confused, ask•If it seems too hard, think some more•Look for hints in other problems•If you are doing math, you’re doing too much•Read directions completely (before & after)•Write down what you know before starting•Draw pictures, define (label) variables•Make sure that unknowns drop out of solution•Don’t forget units!4 P13-FieldsGrass SeedsKnow how to readField LinesKnow how to draw•Field line density tells you field strength•Lines have tension (want to be straight)•Lines are repulsive (want to be far from other lines)•Lines begin and end on sources (charges) or ∞5 P13-PRS Questions:Fields6 P13-E Field and Potential: CreatingA point charge qcreates a field and potential around it:2ˆ;eeqqkVkrr==Er􀁇Use superposition for systems of charges;BBAAVVVVd=−∇∆≡−=−⋅∫EEs􀁇􀁇􀁇They are related:7 P13-E Field and Potential: Creating2ˆ;eeqqkVkrr==Er􀁇Discrete set of point charges:Add up from each point charge2ˆ;eedqdqdkdVkrr==Er􀁇Continuous charge distribution:Break charged object into small pieces, dq, and integrate8 P13-Continuous Sources: Charge DensityCharge Densities:QAσ=QVρ=LQ=λdVdQρ=dLdQλ=dAdQσ=Don’t forget your geometry:2dArdrπ=2cyldVrldrπ=24spheredVrdrπ=dLdx=dLRdθ=9 P13-E Field and Potential: Creating2ˆ;eeqqkVkrr==Er􀁇Discrete set of point charges:Add up from each point charge2ˆ;eedqdqdkdVkrr==Er􀁇Continuous charge distribution:Break charged object into small pieces, dq, and integrateSymmetric charged object:0S;inqdVε⋅=∆≡−⋅∫∫∫EAEs􀁇􀁇􀁇dUse Gauss’ law to get E everywhere, then integrate to get V􀁇􀁷10 P13-0Sinqdε⋅=∫∫EA􀁇􀁇􀁷Gauss’s Law:GaussianPillboxSphericalSymmetryPlanar SymmetryCylindrical Symmetry11 P13-E Field and Potential: Effectsq=FE􀁇􀁇If you put a charged particle, q, in a field:To move a charged particle, q, in a field:WUqV=∆=∆12 P13-PRS Questions:Electric Fields and Potential13 P13-Conductors in EquilibriumConductors are equipotential objects:1) E = 0 inside2) Net charge inside is 03) E perpendicular to surface 4) Excess charge on surface5) Shielding –inside doesn’t “talk”to outside0εσ=E14 P13-PRS Questions:ConductorsP13-CapacitorsTo calculate:1) Put on arbitrary ±Q 2) Calculate E3) Calculate ∆VQCV=∆CapacitanceIn Series & Parallel,parallel12eqCCC=+series12111eq,CCC=+Energy2332oEEudrdrε==∫∫∫∫∫∫2221212VCVQCQU∆=∆==16 P13-PRS Questions:Capacitors17 P13-DielectricsDielectrics locally weakenthe electric field0;1EEκκ=≥0CCκ=Inserted into a capacitor: QCV=∆Hooked to a battery?Q increasesNot hooked up?V decreases18 P13-PRS Questions:Dielectrics19 P13-SAMPLE EXAM:The real exam has 5 concept, 3 analytical questions20 P13-Q: Point ChargesA right isosceles triangle of side 2d has charges q, +2q and -q arranged on its vertices (see sketch). +q2q-qP2d2d(b) What is the potential at P, assuming V(∞)=0?(a) What is the electric field at point P, midway along the line connecting the +q and –q charges?(c) How much work to bring a charge -5Q from ∞to P?21 P13-A: Point ChargesAll charges a distance 2from rdP=+q2q-qP2d2dr3kQr=∑Er􀁇􀁇()3342()()xkkqdEqdqdqdrr=++−−=33;xykkEQxEQyrr→==∑∑(a)()3()2()0ykEqdqdqdr=−++−=()22kQkkqVqqqrrr==+−=∑V=(b)()()()1055kqQWUQVQVPr−=∆=−∆=−=W=(c)22 P13-Q: Ring of ChargeA thin rod with a uniform charge per unit length λ is bent into the shape of a circle of radius Ra)Choose a coordinate system for the rod. Clearly indicate your choice of origin, and axes on the diagram above.b)Choose an infinitesimal charge element dq . Find an expression relating dq , λ, and your choice of length for dq .c)Find the vector components for the contribution of dq to the electric field along an axis perpendicular to the plane of the circle, a distance d above the plane of the circle. The axis passes through the center of the circle. Express the vector components in terms of your choice of unit vectorsd)What is the direction and magnitude of the electric field along the axis that passes through the center of the circle, perpendicular to the plane of the circle, and a distance dabove the plane of the circle.e)What is the potential at that point, assuming V(∞)=0?23 P13-A: Ring of Chargea) Origin & axes as pictureddqdRdλλθ==􀁁b)dqθzxy3kdqdr=Er􀁇􀁇c)()()22ˆˆˆcossin;RRdrRdθθ=−−+=+rijk􀁇d) Horizontal components cancel, only find Ez3zzkdqEdEdr==∫∫230kdRdrπθλθ==∫32kdRrλπ=e) Find the potential by same method:()kdqVddVr==∫∫20kRdrπθλθ==∫2kRrλπ=d24 P13-Q: Spherical CapacitorA conducting solid sphere of radius a, carrying a charge +Qis surrounded by a thin conducting spherical shell (inner radius b) with charge -Q.a) What is the direction and magnitude of the electric field E in the three regions below. Show how you obtain your expressions.1. r< a2. a< r< b3. r> bb)What is the electric potential V(r) in these same three regions. Take the electric potential to be zero at ∞.c)What is the electric potential difference between the outer shell and the inner cylinder, ∆V=V(b) -V(a)?d)What is the capacitance of this spherical capacitor?e)If a positive charge +2Qis placed anywhere on the inner sphere of radius a, what charge appears on the outside surface of the thin spherical shell of inner radius b?25 P13-A: Spherical Capacitora)By symmetry Eis purely radial. Choose spherical Gaussian surface1230Sinqdε⋅=∫∫EA􀁇􀁇􀁷24EAErπ==⋅1&3)00inq=→=E􀁇2)20ˆ4Qrπε=Er􀁇b)For V, always start from where you know it (here, ∞)3) E=0 →Vconstant = 02)()011()4rbQVrdrbπε=−⋅=−∫ES􀁇􀁇1) E=0 →Vconstant = V(a)()0114QVabπε→=−26 P13-()()0114QVbVbaπε⎛⎞=−=−⎜⎟⎝⎠aA: Spherical Capacitorc)123V∆d)()0114QCVabπε−−==∆−e)If you place an additional +2Q charge on the inner sphere then you will induce an additional -2Q on the inner surface of the outer shell, and hence a +2Q charge on the outer surface of that shellAnswer: +2Q27 P13-Q: Find E from VThe graph shows the variation of an electric potential V with distance z . The potential V does not depend on x or y. The potential V in the region -1 m < z < 1 m is given in Volts by the expression V(z)= 15 -5z2. Outside of this region, the electric potential varies linearly with z, as indicated in the graph.(a)Find an equation for the z-component of the electric field, Ez, in the region -1 m < z < 1 m.(b)What is Ezin the region z > 1 m? Be careful to indicate the sign (c)What is Ezin the region z < -1 m? Be careful to indicate the sign (d)This potential is due a slab of charge with constant charge per unit volume ρo. Where is this slab of charge located (give the z-coordinates that bound the slab)? What is the charge density ρoof the slab in C/m3? Be sure to give clearly both the sign and magnitude of ρo.28 P13-A: Find E from V2()155Vzz=−(a)10zVEzz∂=−=∂(b) (z > 1 m)10VmzVEz∂=−=∂These make sense –the electric field points down the hill(c) (z < -1 m)10VmzVEz∂=−=−∂29 P13-A: Find E from V(d)Field constant outside slab, so slab from -1m to 1mThe slab is positively charged since E points away0Sinqdε⋅=∫∫EA􀁇􀁇􀁷2RtLtEAEAEA=+=0in0000Volume2inqAdEAρρεεε===002EAAdερ=GaussianPillboxd=2m0032(10Vm)C10(2m)mεε⎡⎤==⎢⎥⎣⎦30 P13-Q: Parallel Plate CapacitorA parallel plate capacitor consists of two conducting plates of area A, separated by a distance d, with charge +Qplaced on the upper plate and –Q on the lower plate. The z-axis is defined as pictured.a) What is the direction and magnitude of the electric field E in each of the following regions of space: above & below the plates, in the plates and in between the plates.b)What is the electric potential V(z) in these same five regions. Take the electric potential to be zero at z=0 (the lower surface of the top plate).c)What is the electric potential difference between the upper and lower plate, ∆V=V(0) -V(d)?d)What is the capacitance of this capacitor?e)If this capacitor is now submerged into a vat of liquid dielectric (of dielectric constant κ), what now is the potential V(z) everywhere?31 P13-A: Parallel Plate Capacitor(a) Charges are attracted, so live on inner surface only0Sinqdε⋅=∫∫EA􀁇􀁇􀁷Conductors have E=0 inside, and by Gauss’s law the only place E≠0 is between the plates:00downQEAσεε==()0GaussGaussAEAσε=Note that you only need to consider one plate –the other plate was already used (±Q to inner surfaces)32 P13-A: Parallel Plate Capacitor(b)Start where potential is known V(z= 0) = 00()zinVzVd=∆=−⋅∫ES􀁇􀁇0QzAε=−Ez=−Between plates:Above and inside the top conductor E = 0 so Vis constant →V = 00belowQVdAε=−In the bottom plate and below (E=0):33 P13-A: Parallel Plate CapacitordAVQC0ε=∆=(c)()()00QdVVdAε=−=(d)V∆(e) The dielectric constant is now everywhere κ.This reduces the electric field & potential by 1/κ0()inQVzzAκε=−V above and inside top conductor still 00()belowQVzdAκε=−