Engr302 - Lecture 2
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Coulomb’s Law • Coulomb’s Law • Electric Field • Electric Field from Multiple Charges • Integration of Volume charge • Electric Field near Infinite Wire • Electric Field near Infinite Sheet • Electric Field between two Infinite Sheets • Field Lines • Streamlines Coulomb’s Law • Coulomb’s Law with k = 9 x 109 Nm2/C2 εo= 8.85 x 10-12 C2 / Nm2 π1 π2 π1 π2 πΉ=π 2 = π 4πππ π 2 • Unit vector from r1 to r2 πππ ππ − ππ = ππ − ππ • Combining πΉ= π1 π2 4πππ π 12 2 πππ = • (Action reaction F1 = -F2) π1 π2 4πππ π 12 2 ππ − ππ ππ − ππ Example of Coulomb’s Law • Force of charge 1 on charge 2 – Charge 1 - 3 x 10-4 C at M(1,2,3) – Charge 2 - -1 x 10-4 C at N(2,0,5) • Coulomb’s Law πΉ2 = π1 π2 4πππ π 12 2 ππ − ππ ππ − ππ • R magnitude π 12 = ππ − ππ = 2 − 1 ππ + 0 − 2 ππ + 5 − 3 ππ = 3 • Unit vector πππ ππ − 2ππ + πππ ππ − ππ = = ππ − ππ π • Result ππ − 2ππ + πππ 3 × 10−4 πΆ −1 × 10−4 πΆ πΉ2 = = −10ππ + 20ππ − 20ππ π΅ 4π 8.85 × 10−12 πΆ 2 ππ2 (3 π)2 π Electric Field • Electric Field – Coulomb’s Law without 2nd charge – Separates Problem into “Background” and “Test Charge” – Units newtons/coulomb (volts/meter) π¬= ππ π π π − π′ = π = ππ‘ 4πππ π 2 πΉ 4πππ π 2 π − π′ – For source charge at r’ observed at r’ π¬ π = π 4πππ π − π′ 2 π₯ − π₯′ ππ + π¦ − π¦′ ππ + π§ − π§′ ππ π − π′ = 3 π − π′ 4πππ π₯ − π₯′ 2 + π¦ − π¦′ 2 + π§ − π§′ 2 2 • For source charges at r1and r2, observed at r π¬ π = π1 4πππ π − ππ 2 π − ππ π − ππ + π2 4πππ π − ππ 2 π − ππ π − ππ Electric Field from Multiple Charges • 2 source charges at 1 and 2, observed at r π¬ π = π1 π2 π + π 4πππ π − ππ 2 π 4πππ π − ππ 2 π • Multiple source charges at m, observed at r π π¬ π = π=1 ππ π 4πππ π − ππ 2 π • Infinite # source charges, observed at r π¬ π = ππ£ π′ ππ£′ π′ 4πππ π − π′ 2 • We’re going to spend some time on the last one! Example – Electric Field from 4 charges • Sources charges at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0), P4(1,-1,0). Each 3 nC. • Observation point r at P(1,1,1) – P1(1,1,0) π − ππ = π – P2(-1,1,0) π − ππ = 5 – P3(-1,-1,0) π − ππ = π – P4(1,-1,0) π − ππ = 5 π−ππ π−ππ π−ππ π−ππ π−ππ π−π = = π−ππ π−π = π1π πππ +ππ 5 πππ +πππ +ππ = π πππ +ππ 5 • Total field is: 3 × 10−9 πΆ ππ πππ + ππ πππ + πππ + ππ πππ + ππ π¬= + + + 4π 8.85 × 10−12 1 β 1 9β3 5β 5 5β 5 πΈ = 6.82ππ + 6.82ππ + 32.8ππ Continuous Charge - Integration of Charge • Differential charge element ππ = ππ ππ • Integrate for total charge π= π • Example ππ ππ 5 – charge density ππ = −5 × 10−6 π −10 ππ§ πΆ/π2 – Find total charge over region 0 <ρ<1 cm, 2cm <z< 4cm • Comments – Dependence on ρ and z in negative exponential causes rapid fall-off in ρV – Concentrated near z= 0 plane where exponential is small – Concentrated near ρ = 0 z-axis where exponential is small • Integral 0.04 2π 0.01 5 ππ§ −5 × 10−6 π −10 π= 0.02 0 0 π ππ ππ ππ§ Integration of Charge (cont) • Integration on φ 0.04 0.01 5 ππ§ −10−5 π π −10 π= π ππ ππ§ (ππ’ππ‘ππππ¦ ππ¦ 2π) 0.02 0 • Integration on z 0.01 π= 0 −10−5 −105 ππ§ π π −105 π .04 π ππ .02 0.01 π = 10−10 π (πππ£πππ ππ¦ − 105 π ππππππ) π −4000π − π −2000π ππ 0 π ≈ −10−10 π π −4000π π −2000π −10 = 10 π − −4000 −2000 1 1 − −4000 −2000 π = −10−10 π .01 0 (π ππ‘π‘πππ π −2000∗.01 ≈ 0) 1 1 π − = −10−12 = −0.0785 ππΆ 2000 4000 40 Continuous Charge - Other examples • Setup Cartesian 1 – 0.1 ≤ π₯ , π¦ , π§ ≤ 0.2, ππ = π₯ 3 π¦3π§ 3 – Integrate volume −0.2 π‘π + .02 – Subtract volume −0.1 π‘π + .01 – Q will be zero from integration of odd function. • Setup cylindrical – 0 ≤ π ≤ 0.1, 0 ≤ π ≤ π, 2 ≤ π§ ≤ 4, ππ = π2 π§ 2 sin(0.6π) – Differential volume π ππ ππ ππ§ • Universe – ππ = π −2π /π 2 2π π ∞ π= ∞ π 0 0 0 −2π 2 2 /π π π πππ ππππππ = 2π β 2 π 0 −2π 4ππ 0 ππ = − = 6.28 πΆ −2 Continuous Charge -Middle Example • Integral is 4 π 0.1 π2 π§ 2 sin 0.6π π ππ ππ ππ§ π= 2 π= 0 0 .1 π4 4 π = 2.5 0 π§3 3 × 10−5 π = 2.5 × 10−5 4 π sin 0.6π ππ 2 0 56 3 π sin 0.6π ππ 0 56 cos(0.6π) − 3 0.6 π = 1.018 ππΆ 0 Continuous Charge -Field near infinite line charge • Will do in cylindrical coordinates – Observation on y axis, z = 0 plane π = yππ = πππ – Source distributed along z axis π′ = π′ππ – Linear charge density constant ππΏ • Source to observation vector π = π − π′ = πππ − π′ππ • Differential Field Contribution π π¬ = π π¬ = ππΏ ππ§ ′ (π−π′ ) 4πππ π−π′ 3 ππΏ ππ§ ′ (πππ −π§ ′ ππ ) 4πππ π2 +π§ ′2 3 Field near infinite line of charge (cont) • ρ and z components ππΈπ = ππΈπ§ = ππΏ ππ§ ′ (πππ ) 4πππ π2 +π§ ′2 3 2 −ππΏ ππ§ ′ (π§ ′ ππ ) 4πππ π2 +π§ ′2 3 2 (odd - integrates to zero) • Integration for a long wire is thus ∞ πΈπ = ππΏ πππ§ ′ 2 + π§ ′2 4ππ π π −∞ 3 2 ππΏ 1 π§′ πΈπ = π 2 4πππ π π2 + π§′2 ∞ −∞ ππΏ ππΏ = 1 − −1 = 4πππ π 2πππ π ππΏ π¬= π 2πππ π π Field near infinite sheet of charge • Given an infinite line charge and surface density ρs πΈπ = ππΏ 2πππ π • x and y components ππΈπ₯ = ππΈπ¦ = ππ ππ¦′ 2πππ π₯2 ππ ππ¦′ 2πππ π₯ 2 +π¦ 2 + π¦2 πππ π π πππ (odd - integrates to zero) ππΈπ§ = 0 (symmetry) • x component ππΈπ₯ = ππ ππ¦′ 2πππ π₯ 2 +π¦ 2 π₯ π₯ 2 +π¦ 2 = ππ π₯ ππ¦′ 2πππ π₯ 2 +π¦ 2 Field near infinite sheet (cont) • Integration for a sheet is thus ∞ π¬= −∞ ππ π₯ππ¦′ π 2πππ π₯ 2 + π¦ ′2 π ππ π¦′ π¬= π‘ππ−1 2πππ π₯ ∞ ππ −∞ π¬= ππ π −π − ππ 2πππ 2 2 π¬= ππ π 2ππ π • Field points away +ππ +π , toward −ππ −π • Field is independent of distance r<<width Electric Field between 2 Infinite Sheets (-Q) charge sign and unit vector reversed) π¬= ππ −ππ ππ + −ππ 2ππ 2ππ = ππ ππ ππ Field Lines • Field lines – Point in direction of electric field – Direction + test charge moves – Originates on +Q terminates on -Q – Cross-sectional density proportional to E magnitude Streamlines • Equation of line which follows field line at x, y, z – slope of this line y=f(x) – should equal field ratio – Set ππ¦ ππ₯ = ππ¦ ππ₯ πΈπ¦ πΈπ₯ πΈπ¦ πΈπ₯ – Solve for equation y=f(x) as function of x Streamlines • Vector field are Ax, Ay, and Az function of x,y,z • From geometry ππ¦ ππ₯ • Example πΈ= • Plugging in ππ¦ ππ₯ = πΈπ¦ πΈπ₯ π₯ π π₯ 2 +π¦ 2 π = πΈπ¦ πΈπ₯ = • Result πππ¦ = πππ₯ + πππΆ π¦ = πΆπ₯ • Plug in x and y at particular point to evaluate C + π¦ π₯ π¦ π π₯ 2 +π¦ 2 π ππ¦ π¦ = ππ₯ π₯ Streamline Example • Find streamlines of following in rectangular coordinates π¬= • Transforming to rectangular π¬= π¬= 1 ππ β ππ ππ + π₯2 + π¦2 1 π₯2 + π¦2 π¬= • Solution = πΈπ¦ πΈπ₯ = π¦ π₯ At P(-2,7,10) ππ¦ π¦ = πππ¦ = πππ₯ + πππΆ π¦ = πΆπ₯ • πππ π ππ + π₯2 Plugging in streamline equation ππ¦ ππ₯ • 1 π π π y = -3.5 x ππ₯ π₯ 1 ππ β ππ ππ π₯2 + π¦2 1 π₯2 + π¦2 π ππ(π) ππ π₯ π¦ ππ + 2 π 2 +π¦ π₯ + π¦2 π Example problem 1 1. 3 point charges are in xy plane; with 5 nC at y= 5 cm, -10 nC at y =-5 cm, and 15 nC at x=-5cm. Find position of 20 nC that exactly cancels field at origin. - Add first 3 fields to get resultant as function of ax , ay (like example 2.2) - 4th charge must exactly cancel field with same combination of ax , ay - Write in general field form as magnitude times unit vector - Equate magnitudes Example problem 2 7. A 2uC charge is located at A(4,3,5) in free space. Find Eρ, Eφ, and Ez at P(8,12,2) - Get field in rectangular coordinates as function of ax, ay, az - translate rectangular variables to cylindrical variables - translate rectangular unit vectors to cylindrical variables. Example problem 3 13. A uniform charge density extends throughout a spherical shell from r=3 cm to r=5 cm. Find the total charge and the radius containing half the charge. Example problem 4 • Find the electric field on the z-axis produced byan annular ring z= 0, a <ρ <b, 0 < φ < 2π
