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Electromagnetism Physics 15b Lecture #3 Gauss’s Law Electric Potential Purcell 1.13–2.9 What We Did Last Time Defined electric field by F = qE Can be expressed by field lines Defined flux of electric field Φ = ∫ E ⋅ da Gauss’s Law ∫ E ⋅ da = 4π q R Useful for solving E fields with symmetries Spherical charge distribution E ⎧ ⎪ ⎪ E=⎨ ⎪ ⎪⎩ Q r̂ for r ≥ R r2 Qr r̂ for r < R R3 E S Note the sign convention: positive if coming out r Q R2 R r 1 Today’s Goals Continue with Gauss’s Law Apply to infinite sheets of charge Discuss energy in the electric field Empty space with E has energy? Define electric potential ϕ by line-integrating electric field Closely related to energy Vector calculus connects electric potential to electric field and vice versa Derive differential form of Gauss’s Law Connect electric field and charge density More vector calculus J.C.F. Gauss, 1777-1855 Infinite Sheet of Charge Problem: Calculate the electric field at a distance z from a positively charged infinite plane charge Surface charge density: σ = area Use Gauss again Which surface to use? What symmetry do we have? Consider a cylinder Area A and height 2z z E E field must be vertical How do we know that? 2 Infinite Sheet of Charge Total flux Φtotal = Φtop + Φside + Φbottom Side is parallel to E E ⋅ da = 0 No flux Top and bottom are symmetric Same flux Φ total = 2Φ top = 2AE(z) Charge inside the cylinder is z qinside = Aσ E Using Gauss E(z) = 2πσ Don’t forget the direction! ⎧⎪ +2πσ ẑ for z > 0 E=⎨ ⎪⎩ −2πσ ẑ for z < 0 The result is worth remembering: Infinite sheet of charge produces uniform E field of 2πσ above and below Pair of Charged Sheets Place two oppositely-charged large sheets in parallel Consider an area A of them z E fields from the two sheets overlap and add up Etop Ebottom +σ Between the sheets: E = 4πσ Cancel each other outside −σ Two sheets also attract each other (obviously) Top sheet feels The force on area A of the top sheet is Ebottom = −2πσ ẑ F = σ AEbottom E2 = −2πσ Aẑ = − Aẑ 8π 2 3 Pair of Charged Sheets Imagine we move the top sheet upward by a distance d E2 Etop Ebottom We must do work W = Fd = Ad z 8π The energy of the system increases +σ by W Q: Where exactly is this energy? −σ Note that the volume of the space between the sheets increased by Ad This is also where E field exists E2 Space with E holds energy with a volume density u = 8π Total electrostatic energy of a system is E2 U=∫ dV Will come back to this… 8π Line Integral of Electric Field Electrostatic force is conservative I said this in Lecture 1 without proof Given F = qE, the above statement is equivalent to Line integral ∫ P2 P1 E ⋅ d s is path independent Thanks to the Superposition Principle, we have only to prove this for the electric field generated by a single point charge Line integral ∫ r2 r1 E= q r̂ r2 ds r q r r̂ ⋅ d s is path independent 1 2 r r2 q 4 Line Integral of Electric Field The dot product r̂ ⋅ ds is the radial component of the movement, i.e. dr ∫ r2 r1 q r̂ ⋅ d s = r2 ∫ r2 r1 ⎛ 1 1⎞ q dr = q ⎜ − ⎟ 2 r ⎝ r1 r2 ⎠ E= dr The integral depends only on r1 and r2, i.e., is path-independent ds Generalize using Superposition: Line integral ∫ P2 P1 q r̂ r2 r E ⋅ d s for any electrostatic field E has the same value for all paths from P1 to P2 r1 r2 q Corollary For the special case of P1 = P2, the path becomes a loop Line integral ∫ E ⋅ d s of an electrostatic field around any closed path is zero This is equivalent to the path-independence Consider two paths (A and B) from P1 to P2 Path independence means ∫ P2 P1 ∫ P2 P1 E ⋅ d sB Corollary above means ∫ P2 P1 E ⋅ d sA = P1 dsA P2 dsB P1 E ⋅ d s A + ∫ E ⋅ (− d sB ) = 0 P2 Will use this later when we do the “curl” 5 Electric Potential The line integral P2 ∫ P2 P1 E ⋅ d s is useful enough to have a name φ21 ≡ − ∫ E ⋅ d s Electric potential difference between P1 and P2 P1 Note the negative sign! To move a charge q from P1 to P2, you must do work W= ∫ P2 P1 −qE ⋅ d s = qφ21 As a result, the energy of the system increases by qϕ21 We can fix P1 to a reference point and re-interpret this as a scalar function of the position r r φ (r) ≡ − ∫ E ⋅ d s Electric potential at position r 0 Reference Point Reference point for the electric potential is arbitrary If you choose e.g., point B instead of point A as the reference r r A B A B φref =B (r) = − ∫ E ⋅ d s = − ∫ E ⋅ d s − ∫ E ⋅ d s = φref = A (r) + const. Electric potential is defined up to an arbitrary constant Just like an indefinite integral The potential difference is physical (linked to work and energy) and must be free from arbitrary constant P2 P2 P1 P1 0 0 φ21 = − ∫ E ⋅ d s = − ∫ E ⋅ d s + ∫ E ⋅ d s = φ (P2 ) − φ (P1) The constant cancels in the difference 6 Unit Dimension of electric potential is (electric field)×(length) Since electric field is (force)/(charge), this equals to (force)×(length)/ (charge) = (energy)/(charge) Unit of electric potential is erg/esu = statvolt In SI, the unit of electric potential is volt = joule/coulomb 300 volt = 1 statvolt 1 coulomb = 3×109 esu Since the rest of the world uses SI, we will convert to SI when we deal with real-world (esp. EE) problems Electric Field ↔ Potential Recall in calculus: F(x) = ∫ f (x) dx f (x) = We can “reverse” the line integral as well φ = −∫ E ⋅ds E = −∇φ d F(x) dx gradient Gradient is “how quickly the function ϕ varies in space” In x-y-z coordinate system: ∇φ = x̂ ∂φ ∂φ ∂φ + ŷ + ẑ ∂x ∂y ∂z NB: gradient is a vector field It points the direction of maximum rate of increase (= uphill) Hence E points downhill of ϕ 7 Visualizing Gradient in 2-d For a potential φ (x,y) = sin x ⋅ sin y ∇φ (x,y) = x̂ cos(x) sin(y) + ŷ sin x cos y ∇φ (x,y) φ (x,y) © Prof. G. Sciolla, MIT Equipotential Surfaces The same potential φ (x,y) = sin x ⋅ sin y can be expressed with lines of constant values of ϕ In 3-d, we find surfaces of constant potential = equipotential surfaces For a single point charge, equipotential surfaces are concentric spheres ∇φ (x,y) Electric field is perpendicular to the equipotential surfaces This is generally true for gradient of any function © Prof. G. Sciolla, MIT 8 Charge Distribution Electric potential due to a charge distribution is N φ=∑ j =1 qj φ= discrete rj ∫ dq r continuous Continuous case maybe 1-d, 2-d, or 3-d φ= λ ∫ r d φ= σ ∫r da φ= ρ ∫ r dv Since potential is a scalar, it is often easier to calculate than the electric field Once you have ϕ, you can always get E from −∇ϕ An example is in order Charged Disk Problem: Calculate the electric potential produced by a thin, uniformly charged disk on its axis z Disk radius = a, surface charge density σ Slice the disk into rings, then into the red bits Charge on the red bit is σsdsdθ σ sdsdθ The potential due to the red bit is dφ = z2 + s 2 Integrate: 2π a sds 0 0 z +s φ = σ ∫ dθ ∫ 2 ds 2 dθ s s =a = 2πσ ⎡⎢ z 2 + s 2 ⎤⎥ ⎣ ⎦s =0 = 2πσ ( z +a − z 2 2 ) a NB: absolute value in case z < 0 9 Electric Field Can we calculate E from φ = 2πσ ( ) z 2 + a2 − z ? We need to take gradient, which means we need to know the x-, y-, and z-dependence z We calculated ϕ only on the z axis We are in luck — we know E is parallel to the z axis by symmetry E = −∇φ = − ẑ = −2πσ ∂ ∂z ( ∂φ ∂z ds ) dθ s z 2 + a 2 − z ẑ ⎛ ⎞ z = 2πσ ⎜ − ± 1⎟ ẑ ⎝ ⎠ z 2 + a2 a for z > 0 z<0 Check the Solution φ = 2πσ ( z 2 + a2 − z ) ⎛ ⎞ z E = 2πσ ⎜ − ± 1⎟ ẑ ⎝ ⎠ z 2 + a2 for z > 0 z<0 Dimension: [ϕ] = (charge)/(length), [E] = (charge)/(length)2 NB: [σ] = (charge)/(length)2 Far away from the disk (|z| >> a) φ = 2πσ z ( 1+ (a z) − 1) ⎯⎯⎯→ 2πσ z (1+ z a 2 1 2 ) (a z)2 − 1 = π a 2σ z Same as a point charge with Q = πa2σ Close to the disk (|z| << a) z a E ⎯⎯⎯ → ±2πσ ẑ Same as the infinite sheet of charge 10 Energy and Potential We know a system of N charges has a total energy of q j qk 1 N q 1 N 1 N Potential at the j-th U = ∑∑ = ∑ q j ∑ k = ∑ q j ∑ φ jk 2 j =1 k ≠ j r jk 2 j =1 k ≠ j r jk 2 j =1 k ≠ j charge due to the other charges Generalizing to continuous distributions U= 1 ρφ dv 2∫ Integrate entire space, or where ρ ≠ 0 No need to avoid j = k because “individual” charge is infinitesimally small Does the factor 1/2 make sense? Imagine increasing ρ slowly everywhere as ρ ′ = s ρ s = 0 → 1 The potential is proportional to ρ, i.e., φ ′ = sφ Work to go from s to s + ds is dW = (d ρ ′)φ ′ dv = (ds ρ )(sφ ) dv ∫ 1 Integrate W = ∫0 s ds ∫ ρφ dv = 2 ∫ ρφ dv 1 ∫ Will come back to this… Shrinking Gauss’s Law Charge is distributed with a volume density ρ(r) Draw a surface S enclosing a volume V Guass’s Law: ∫ S E ⋅ da = 4π ∫ ρ dv Total charge in V V Now, make V so small that ρ is constant inside V ∫ S E ⋅ da = 4πρV for very small V As we make V smaller, the total flux out of S scales with V Therefore: ∫ lim S V →0 E ⋅ da V = 4πρ LHS is “how much E is flowing out per unit volume” Let’s call it the divergence of E 11 Divergence div E ≡ lim V →0 ∫ S E ⋅ da V = 4πρ In the small-V limit, the integral depend on volume, but not on the shape We can use a rectangular box Consider the left (S1) and right (S2) walls ∫ ∫ S1 S2 dz E ⋅ da = E(x,y,z) ⋅ (− x̂)dydz E ⋅ da = E(x + dx,y,z) ⋅ x̂dydz ( ) dy Sum = E x (x + dx) − E x (x) dydz = ∂E x dxdydz ∂x Add up all walls: S1 E(x,y,z) E(x + dx,y,z) S2 dx ⎛ ∂E ∂E y ∂E z ⎞ E ⋅ da = ⎜ x + + ⎟ V = ∇ ⋅E V S ∂y ∂z ⎠ ⎝ ∂x ( ∫ ) div E Gauss’s Law, Local Version We now have Gauss’s Law for a very small volume/surface divE = 4πρ ⎛ ∂E ∂E y ∂E z ⎞ where div E ≡ ∇ ⋅E = ⎜ x + + ⎟ ∂y ∂z ⎠ ⎝ ∂x Connects local properties of E with the local charge density Integrate over a volume and you get Gauss’s Law back Gauss’s Divergence Theorem (this is math!) For any vector field F, ∫ V div F dv = ∫ A F ⋅ da Applying this theorem to Gauss’s Law (of electromagnetism) gives us the integral and differential versions we now know 12 Coulomb Field Let’s calculate div E for E = q r̂ r2 q x̂x + ŷy + ẑz 2 2 x + y + z x2 + y 2 + z2 Or we can express div in spherical coordinates 2 1 ∂(r Fr ) 1 ∂(Fθ sin θ ) 1 ∂Fφ ∇ ⋅F = 2 + + ∂r r sin θ ∂θ r sin θ ∂φ r We can do this by expressing E in x-y-z : E = Since only Er is non-zero, we get ∇ ⋅E = 2 2 1 ∂(r Er ) 1 ∂q = 2 =0 r 2 ∂r r ∂r This is correct — we have no charge except at r = 0 At r = 0, 1/r2 gives us an infinity That’s OK because a “point” charge has an infinite density Spherical Charge Let’s give the “point” charge a small radius R We did this in Lecture 2, and the solution was ⎧ ⎪ ⎪ E=⎨ ⎪ ⎪⎩ Q r̂ for r ≥ R r2 Qr r̂ for r < R R3 This part is same as a point charge. We know div E = 0. Let’s work on this part 2 1 ∂(r Er ) 1 ∂ ⎛ Qr 3 ⎞ 3Q = 2 = 2 ∂r r r ∂r ⎜⎝ R 3 ⎟⎠ R 3 For r < R, ∇ ⋅E = The charge density of the sphere is ρ = Q R3 4π 3 ∇ ⋅E = 4πρ It works everywhere (as long as ρ is finite) 13 Energy Again We found earlier U = Are they same? 1 E2 ∫ 8π dV and U = 2 ∫ ρφ dv Consider the divergence of the product Eϕ ∂ ∂ ∂ (E φ ) + (E φ ) + (E zφ ) ∂x x ∂y y ∂z ∂E x ∂φ ∂E y ∂φ ∂E z ∂φ = φ + Ex + φ + Ey + φ + Ez ∂x ∂x ∂y ∂y ∂z ∂z ∇ ⋅ (Eφ ) = = (∇ ⋅E)φ + E ⋅ (∇φ ) = 4πρφ − E 2 Integrate LHS over very large volume and use Divergence Theorem ∫ V ∇ ⋅ (Eφ ) dv = ∫ S (Eφ ) ⋅ da = 0 Integral of RHS must be 0, too 4π ∫ ρφ dv − ∫ E 2 dv = 0 assuming Eφ → 0 at far away 1 E2 ρφ dv = ∫ 8π dv 2∫ Summary Used Gauss’s Law on infinite sheet of charge Uniform electric field E = 2πσ above and below the sheet E2 Electric field has energy with volume density given by u = 8π Defined electric potential by line integral P2 φ21 = − ∫ E ⋅ d s = φ (P2 ) − φ (P1) unit: erg/esu = statvolt P1 Electric field is negative gradient of electric potential Potential due to charge distribution: φ = N ∑r j =1 Differential form of Guass’s Law: qj E = −∇φ or φ = j ∫ dq r ∇ ⋅E = 4πρ Equivalent to the integral form via Divergence Theorem ∫ V div F dv = ∫ A F ⋅ da 14

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