PHYS 216 Assignment #4 – Solutions PHYS 216 – Assignment #4 – Solutions – Due Oct 21 Chapters 7–8: Potential, Conductors & Capacitance Throughout, ε0 = 8.854 × 10−12 C2 /N m2 . Problem 1. (a) Take a coaxial Gaussian cylinder of radius s and length L (a < s < R). By ⃗ symmetry E is radial and constant over the curved surface, and there is no flux through the ends. Only the wire’s charge is enclosed: λL λ E (2πsL) = ⇒ E(s) = (radially outward). ε0 2πε0 s (b) Z R Z R λ ds λ R V0 = V (a) − V (R) = E ds = = ln . a 2πε0 a s 2πε0 a (c) From (b), λ/2πε0 = V0 / ln(R/a). Substituting into (a): V0 E(s) = . s ln(R/a) (d) Here ln(R/a) = ln 200 = 5.30, so 900 V E(a) = ≈ 1.7 × 106 V/m, (1.0 × 10−4 m)(5.30) E(1.00 cm) ≈ 1.7 × 104 V/m, E(R) ≈ 8.5 × 103 V/m. Because E ∝ 1/s, the field at the wire is R/a = 200 times larger than at the wall. Only a thin sheath around the wire has a field approaching the breakdown scale (E(a) is more than half of the 3 × 106 V/m needed for dry air; real Geiger tubes use a low-pressure fill gas that breaks down at a lower field). Throughout the rest of the tube the field is two orders of magnitude below breakdown, so ionising avalanches start only next to the wire. For a given V0 , E(a) = V0 /[a ln(R/a)] grows rapidly as a shrinks (the ln varies only slowly). A rod of radius 1 cm would give only E(a) = 900/(0.01 ln 2) ≈ 1.3 × 105 V/m, far below breakdown. A thin wire produces a very large field with a modest, safe voltage. (This is the same physics as charge crowding at sharp points: σ = ε0 E is large on the thin wire.) (e) C λ 2πε0 2π(8.854 × 10−12 ) = = = = 10.5 pF/m . L V0 ln(R/a) 5.30 Problem 2. (a) ε0 A ε0 A C0 = , C= . d d−t With the slab in place the field is zero inside the conductor, leaving two gaps x1 + x2 = d − t in series: 1/C = x1 /ε0 A + x2 /ε0 A = (d − t)/ε0 A. Only the sum of the gaps enters, so the vertical position does not matter. 1 PHYS 216 Assignment #4 – Solutions (b) The capacitor is isolated, so Q is fixed: Q2 d Q2 (d − t) Q2 t U0 = , U= , ∆U = − < 0. 2ε0 A 2ε0 A 2ε0 A The energy decreases: the field Q/ε0 A is unchanged in the gaps but has been removed from the volume At occupied by the slab. (c) The field points up, from the +Q bottom plate to the −Q top plate. Field lines leaving the bottom plate end on the slab’s bottom face, and the lines ending on the top plate start on the slab’s top face: σtop face = +Q/A , σbottom face = −Q/A . The slab’s total charge remains zero. (d) Sketch: uniform vertical lines between the plates and the slab where the slab is in; uniform lines across the full gap where it is not; near the slab’s leading edge the fringing lines bend to end on the slab’s edge and corners. These fringing lines have a horizontal component at the slab’s induced charges, pulling them (and so the slab) toward the region between the plates: the slab is pulled in. Equivalently, let ℓ be the distance the slab has been pushed in (not to be confused with the gaps x1 , x2 ). At fixed Q the energy decreases as ℓ increases, so F = −dU/dℓ points inward. To slide it in slowly you must hold it back, so you do negative work : Q2 t Wyou = ∆U = − . 2ε0 A (e) Now the battery holds the potential difference fixed at its emf V , and U = 12 CV 2 increases because C increases: ε0 AV 2 t ∆U = 12 (C − C0 )V 2 = > 0. 2 d(d − t) The battery moves extra charge ∆Q = (C − C0 )V through potential difference V , doing work Wbatt = ∆Q V = (C − C0 )V 2 = 2 ∆U. Energy conservation, Wbatt + Wyou = ∆U , gives Wyou = −∆U : half the battery’s work goes into stored energy, and the other half is work done by the field on the slab (taken up by your hand). The slab is still pulled in (at fixed V , F = +dU/dℓ = 12 V 2 dC/dℓ, inward), and you still do negative work. The force depends on the local fields near the slab’s edge, not on whether a battery is attached; the sign of ∆U differs only because the battery supplies energy. Problem 3. (a) Ceq = C1 C2 /(C1 + C2 ) = 1.33 µF. In series both carry the same charge: Q = Ceq V = 1.33 mC each, V1 = Q/C1 = 667 V , V2 = Q/C2 = 333 V , U0 = 12 Ceq V 2 = 0.667 J . (b) Joining like plates puts the two positive plates together, so the total charge on that side is 2Q = 2.67 mC, shared between two capacitors in parallel at a common voltage: 2Q Vf = = 444 V across each, C1 + C2 Q1 = C1 Vf = 0.889 mC , Q2 = C2 Vf = 1.78 mC . 2 PHYS 216 Assignment #4 – Solutions (c) U0 − Uf Uf = 21 (C1 + C2 )Vf2 = 0.593 J , = 91 ≈ 11% . U0 (Exactly: U0 = 12 Q2 (1/C1 + 1/C2 ) = 3Q2 /8 µF−1 and Uf = (2Q)2 /2(C1 + C2 ) = Q2 /3 µF−1 , so Uf /U0 = 8/9.) The missing 0.074 J is dissipated as heat in the connecting wires (plus a little electromagnetic radiation) while charge flows to equalise the voltages. A smaller resistance does not reduce the loss: it gives a larger current for a shorter time, but the energy dissipated is set only by the initial and final states. Problem 4. (a) Between the conductors (Gauss) E = Q/4πε0 r2 , so Z b   Q 1 1 Q 4πε0 ab ∆V = E dr = − , C= = . a 4πε0 a b ∆V b−a (b) Q2 Q2   1 1 U= = − . 2C 8πε0 a b (c) Z b 2 Z b Q2 Q2    ε0 Q 2 dr 1 1 U= 4πr dr = = − ,✓ a 2 4πε0 r2 8πε0 a r 2 8πε0 a b in agreement with (b). Q2 (d) As b → ∞, U = (C = 4πε0 a). The energy stored between a and r is proportional to 8πε0 a 1/a − 1/r, so the fraction within r is 1 − a/r: within 2a : 1/2 , 1 − a/r = 0.9 ⇒ r = 10a . Half the energy lies within one radius of the surface, even though the field extends to infinity. (e) Q = 4πε0 aV = 4π(8.854 × 10−12 )(0.15)(3.0 × 105 ) = 5.0 µC , and U = 12 QV = 0.75 J . 3