PHY2049: Chapter 221Chapter 22: Electric FieldPHY2049: Chapter 222Electric Field of Single Point Charge2ˆkqErr=G2ˆkqErr=−GPHY2049: Chapter 223Example: Electric Field on ProtonÎAt surface of proton q = e = 1.6 x 10-19C r = 10-15mÎE points radially outward for + charge()()()919212215910 1.6101.44 10 N/C10kqEr−−××== = ×PHY2049: Chapter 224E Field of Two Equal, Positive Point ChargesPHY2049: Chapter 225E Field of Two Equal, Unlike Point ChargesPHY2049: Chapter 226Field Between Two Charged Parallel PlatesÎAssume plates are much larger than separation E is approx. constant between plates E is zero outside the plates This is a capacitor!ÎE points from + plate to – plateÎWe will calculate E in Chap. 23 Gauss’ law Proportional to surface charge densityPHY2049: Chapter 2271. Rank magnitude of E at P1, P2, P3.Assume charges on rings are +q and +qPHY2049: Chapter 228Answer to Question #1ÎP1has E = 0 since it is equidistant from ring A and B and they are same sign.ÎP3has largest E because it has contributions from ring A and BÎP2has no contribution from ring B because it is at the center, thus it is only affected by ring A.ÎSo the order (smallest E to largest E) is P1, P2, P3PHY2049: Chapter 2292. Rank magnitude of E at P1, P2, P3.Assume charges on rings are +q and −qPHY2049: Chapter 2210Answer to Question #2ÎP1has largest E field since it is equidistant from ring A and B and their E contributions add, rather than cancel, as in the first question.ÎHard to rank E field of P3and P2, in my opinion. Relative distances from the two rings are different and there is a cancellation in P3.PHY2049: Chapter 2211Calculate E of Dipole (⊥ axis)ÎAt point x, Ex= 0 and Ey< 0. Why?+Q-Qd()222 3/222222/2sin 2/4/4/4ykQ kQ d kQdErxdxdxdθ−− −⎛⎞== =⎜⎟+⎝⎠++3ykpExdx≈− (dipole moment)pQd=xrr22/4rxd=+/2sindrθ=θPHY2049: Chapter 2212Calculate E of Dipole (along axis)ÎAt point x, Ex> 0 and Ey= 0. Why?x−Qd()()()22 2222/2 /2/4xkQ kQ kQxdExd xdxd=−=−+−32xkpExdx≈ (dipole moment)pQd=x+QPHY2049: Chapter 2213Finding E Field from Charge DistributionÎPerform integral over charge distribution Each component must be calculated separately (vector addition)ÎGeneral helpful rules Use symmetry to see if any component must be zero Use symmetry to see if any component is doubled, etc. Express dq, r and trig functions in terms of “natural” variables defined by the problem Then we can integrate!()2sin or cosykdqdErθθ=PHY2049: Chapter 2214θdqCenter of Uniformly Charged CircleÎE field is down. Why? Uniform distribution of charge Express dq, r, sinθ in terms of θ Top, bottom give same contribution()22sin sinykrdkdqdErrλθθθ=− =−()2002sin2cos4ykrdErkrkrππλθθλθλ=× −−=−−=∫qrλπ=24ykqErπ=−dq ds rdλλθ==PHY2049: Chapter 2215Axis of Uniform Charged Ring (+Q)ÎPoint is distance z above center of charged ring, radius R Uniformly charged (ind. of angle φ) Horizontal components cancel()222sinsinzkRdkdqdErzRλφθθ==+2QRλπ=()()()23/2 3/2022 223/2222zzkzRd kz REzR zRkQzEzRπλφπλ==++=+∫zdq Rdλφ=Rx22sinzzrzRθ==+θCheck this!PHY2049: Chapter 2216Behavior at Large and Small zÎExactÎFor z smallÎFor z large()3/222zkQzEzR=+3zkQzER→2zkQEz→Coulomb’s lawAs expectedBecomes 0 when z = 0PHY2049: Chapter 2217Uniformly Charged LineÎDistance y above midpoint of charged line of length L Uniformly chargedExcomponents cancel()()()/2 /2222 3/2/2 /2222222sin/4LLyLLykdx kydxkdq yEryxyxyxkLEyy Lλλθλ−−== =+++=+∫∫ ∫/QLλ=yPxLdq dxλ=θCheck this!PHY2049: Chapter 2218Special Cases for Charged LineÎ Exact expressionÎInfinite line (L →∞)Î Zero length (point)22/4ykLEyy Lλ=+2ykEyλ=22ykL kQEyyλ==Coulomb’s lawAs expectedFalls as inverse distancePHY2049: Chapter 2219Uniformly Charged DiskÎPoint is distance z above axis of charged disk, radius R()()()222200223/2022sin2zRRkdqErkddzzzkzdzπθσρ ρ φρρπσρ ρρ==++=+∫∫∫∫2QRσπ=z()dq dAddσσρρφ==Rθρ2221zzEkzRπσ⎛⎞=−⎜⎟⎜⎟+⎝⎠PHY2049: Chapter 2220Charged Disk (cont)ÎExact expression Îz << RÎz >> R022zEkσπσε→=2221zzEkzRπσ⎛⎞=−⎜⎟⎜⎟+⎝⎠Independent of z!See next chapter222zkR kQEzzπσ→=Coulomb’s lawAs
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