lect03 - • Physics 227: Lecture 3 Electric Field...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: • Physics 227: Lecture 3 Electric Field Calculations, Symmetry, Charged Particle Motion, Flux, Gauss’s Law Lecture 2 review: If you know anyone who has not managed to register, some more spaces have opened up. Please have them r2 contact Stacey Jacbs ASAP. • • F1 on 2 = k q1 q2 / • Electric field determined by test charge: E = Fon qtest / qtest • Superposition: Forces are vectors, Coulomb forces add as ``Cheat’’ / formula sheet added vectors. to Sakai resources - tell me about any typos • Electric fields are vectors, and add as vectors. Unique direction and magnitude at each point in space. • Efrom point charge = k q / r2 Field lines show direction of electric field: E tangent to find line, fields lines start on + charge, end on - charge, but may start / stop at r = ∞, field strength ∝ number of field lines, number of field lines ∝ charge. Monday, September 12, 2011 Calculating the Electric Field • Superposition allows the electric field to be calculated for an arbitrary fixed charge distribution ρ(x,y,z). Let: ￿x E (￿ ) = ￿ • ￿ i Or: ￿x E (￿ ) = ￿ qi ˆ 2 ri → 4π￿0 ri ￿ i qi ˆ 2 ri → 4π￿0 ri ￿ ρ(￿ ) x ˆ ∆ dx 4π￿0 |￿ − ￿ ￿ |2 xx ￿ 1ˆ dq ∆ 4π￿0 r2 3 • Where Δ is the unit vector from the position of the charge to the position at which we are evaluating the electric field. • The integrals might require numerical evaluation, turning this into a computation problem. Monday, September 12, 2011 Electric Field of Two Charges y q1 q2 d ￿x E (￿ ) = ￿ ￿ i • • • qi ˆ 2 ri → 4π￿0 ri x d ￿ ρ(￿ ) x ˆ ∆ dx 4π￿0 |￿ − ￿ ￿ |2 xx 3 What is the electric field along the x axis? For x > d: For 0 < x < d: Monday, September 12, 2011 q2 q1 ￿ x+ ˆ x ˆ E= 4π￿0 |x − d|2 4π￿0 |x + d|2 −q1 q2 ￿ E= x+ ˆ x ˆ 2 2 4π￿0 |x − d| 4π￿0 |x + d| Notice that at x >> d the field falls like the field of a point charge Electric Field of Two Charges y ￿x E (￿ ) = ￿ E ￿ i -q1 q1 d d qi ˆ 2 ri → 4π￿0 ri ￿ ρ(￿ ) x ˆ ∆ dx 4π￿0 |￿ − ￿ ￿ |2 xx 3 • x What is the electric field along the y axis? • You can see the vertical components will cancel. −2q1 d ￿ E= x× ￿ ˆ 4π￿0 (y 2 + d2 ) y 2 + d2 −2q1 d ￿ E= x ˆ 4π￿0 (y 2 + d2 )3/2 Monday, September 12, 2011 Notice that at large y the dipole field falls faster than the field of a point charge Electric Field of Two Charges • Why y does the field fall faster than the field of a point charge? E A. It cannot - the professor is wrong! -q1 q1 d d B. It always falls faster for 2 or more charges. x C. The two charges cancel, so the field is 0 everywhere! D. Because the fields of the two charges partially cancel. E. It falls faster in some directions, but not all. ￿ E= Monday, September 12, 2011 Notice that at large y −2q1 d the dipole field falls x faster than the field ˆ 4π￿0 (y 2 + d2 )3/2 of a point charge Electric Field of Two Charges • Why y does the field fall faster than the field of a point charge? E A. It cannot - the professor is wrong! -q1 q1 d d B. It always falls faster for 2 or more charges. x C. The two charges cancel, so the field is 0 everywhere! A. Often, but not this time. B. Not for, e.g., two same D. Because the fields of the two charges partially cancel. sign charges. C. There is a partial, not E. It falls faster in some directions, but not all. total, cancellation. E. Dipole field falls like 1/r3, Notice that at large y −2q1 d though you have not been ￿ the dipole field falls E= x faster than the field ˆ told enough so far to know 4π￿0 (y 2 + d2 )3/2 of a point charge this answer is wrong Monday, September 12, 2011 Electric Field of a Ring of Charge Example 21.10 • • ￿x E (￿ ￿ ) = ￿ i qi ˆ 2 ri → 4π￿0 ri 1ˆ dq ∆ 4π￿0 r2 You can see the vertical components will cancel. • ￿ What is the electric field along the x axis? The integrand does not depend on position along the ring, so the integral can be done ``by inspection’’ Notice that at large x >> a, this approaches the field of a point charge. Monday, September 12, 2011 Symmetry Arguments Sometimes you can determine something about the shape of the electric field from symmetry arguments For a point charge or a uniform sphere of charge, the field is radial and spherically symmetric: E(r,θ,φ) → E(r) + + + The charge looks the same to both observers. The charge looks the same when an observer rotates about its head. There is nothing to distinguish the observer’s up from your down, or left from right. There is no preferred direction in space except for the line going from the charge to the observer. The field must be along this line. (You are a point in space, your body does not count!) Monday, September 12, 2011 Symmetry Arguments Sometimes you can determine something about the shape of the electric field from symmetry arguments For an infinitely long uniform line of charge, the field is radial and perpendicular to the line: E(ρ,φ,z) → E(ρ) + + + + + + + + + + + + φ + z ρ If you move in z, the line looks the same, so the field has no z component. If you go around the line in the φ direction, the line looks the same, so the field has no φ component. The field can only be in the ρ direction. Monday, September 12, 2011 Symmetry Arguments Sometimes you can determine something about the shape of the electric field from symmetry arguments For an infinite uniform plane of charge, the field is perpendicular to the plane: E(x,y,z) → E(z) + + + + + + + + + + + + + + y + + + + + If you move in x or y, or if you rotate about the z axis, the plane looks the same, so the field has no x or y components. The field can only be in the z direction. Monday, September 12, 2011 x Motion in an Electric Field • Soon we will learn that the field inside a ``parallel plate capacitor’’ is ≈ constant, as long as the size of the plates (√A) is much greater than the separation of the plates (d) x • + v0 y Since the E-field is constant, the electric force F = qE is constant between the plates. • Thus, F = ma leads to ay = qE/m and ax = 0. • With v(t=0) = v0x, we find that x = v0xt, and y = (1/2)at2 = (1/2)(qE/m)t2. • Put t = x/v0x into the equation for y to obtain: y = (1/2)(qE/mv0x2)x2. • This is a parabolic trajectory, just like motion of a projectile in a uniform gravitational field. It does not follow the field lines. Monday, September 12, 2011 Introduction to Flux • For a constant E-field, flux through a surface is, crudely speaking, the number of field lines through the surface. • Toy problem: inside the parallel plate capacitor the E field is constant and vertical, 25 field lines per m2 • There is a 1 m2 horizontal surface in side the capacitor. • What is the flux through this horizontal surface? • Flux = field line density x area = 25 field lines/m2 x 1 m2 Monday, September 12, 2011 = 25 field lines Toy Examples, with flux of water Velocity of water is a vector field, like electric fields. • Water of density 1000 kg/m3 flows at 3 m/s through a horizontal pipe of cross sectional area 1 m2. What is the flux of water through a vertical plane through the pipe? • Flux = 1000 kg /m3 x 3 m/s x 1 m2 = 3000 kg/s. Water of density 1000 kg/m3 flows at 3 m/s through a horizontal • pipe of cross sectional area 1 m2. What is the flux of water through a horizontal plane through the pipe? • Flux = 0. No water goes through the plane! Water flows along the plane, but it does not cross over it. Monday, September 12, 2011 More toy examples • Water of density 1000 kg/m3 flows at 3 m/s through a U-shaped horizontal pipe of cross sectional area 1 m2. What is the flux of water through a vertical plane through the upper+lower pipes? • • • Fluxupper = 1000 kg /m3 x 3 m/s x 1 m2 = 3000 kg/s. Fluxlower = 1000 kg /m3 x -3 m/s x 1 m2 = -3000 kg/s. Fluxtotal = Fluxupper + Fluxlower = 0 Monday, September 12, 2011 More formal definition of Flux E-field lines ΦE = ˆ Surface: A unit vector ⊥ to surface try ΦE = ￿ ￿ ￿ ￿ E · dA = ˆˆ A=x ￿ E cos φdA = E⊥ dA ￿ E = E0 (cos φx + sin φy ) ˆ ˆ E0 (cos φx + sin φy ) · dA x = E0 cos φA ˆ ˆ ˆ More generally we have a curved surface with ￿ E = Ex (x, y, z )ˆ + Ey (x, y, z )ˆ + Ez (x, y, z )ˆ x y z and the integral will be hard to evaluate. Monday, September 12, 2011 ￿ Closed Surfaces + Gauss’s Law • We are going to be concerned w ith the total electric flux through a closed surface. • In advanced integral calculus E&M, you learn that the surface integral of the flux can be related to a volume integral that depends on the charge in the volume. • The book does not show this; we will just accept the result: ΦE = ￿ ￿ · dA = qenclosed E￿ ￿0 The result does not depend on the shape of the closed surface! Monday, September 12, 2011 Closed Surfaces • There is a charge distribution near the origin. • The charge furthest from the origin is a distance a away. • There is a Gaussian sphere of radius b centered at the origin, with b > a. •In which of the following cases does the flux through the surface definitely not change? •There may be more than one right answer. A. The radius of the sphere decreases to a/2. B. The magnitude of each charge is doubled. C. The sphere moves sideways so its surface goes through the origin. D. The sphere deforms into a cube with each side 2a long centered at the origin. E. The charge doubles in magnitude and the sphere doubles in surface area to compensate. Monday, September 12, 2011 Closed Surfaces • There is a charge distribution near the origin. • The charge furthest from the origin is a distance a away. • There is a Gaussian sphere of radius b centered at the origin, with b > a. •In which of the following cases does the flux through the surface definitely not change? •There may be more than one right answer. A. The radius of the sphere decreases to a/2. A.,B.,C.: charge would usually change, but might B. The magnitude of each charge is doubled. not in special cases C. The sphere moves sideways so its surface goes through the origin. D. The sphere deforms into a cube with each side 2a long centered at the origin. E. The charge doubles in magnitude and the sphere doubles in surface E. charge definitely changes area to compensate. Monday, September 12, 2011 Flux through Closed Surfaces Monday, September 12, 2011 Closed Surfaces • • Six charges are in a plane. For simplicity here, treat all charges as +/-1 μC, ignore the magnitudes shown. • The intersections of 5 closed surfaces with the plane are shown. A. S1 B. S2 • Through which surface is the flux greatest? •There may be more than one right answer C. S3 D. S4 E. S5 Monday, September 12, 2011 Closed Surfaces • • Six charges are in a plane. For simplicity here, treat all charges as +/-1 μC, ignore the magnitudes shown. • The intersections of 5 closed surfaces with the plane are shown. A. S1 • Through which surface is the flux greatest? • There may be more than one right answer. • q in S1 = 0, q in S2 = +1, q in S3 = +2, q in S4 0, q in S5 = +2 (all in μC). B. S2 = C. S3 D. S4 E. S5 Monday, September 12, 2011 See you Thursday Monday, September 12, 2011 ...
View Full Document

This note was uploaded on 09/27/2011 for the course PHYSICS 750:227 taught by Professor Ronaldgilman during the Fall '11 term at Rutgers.

Ask a homework question - tutors are online