TUTORIAL 4. Proof. Computing the potential at the center and pole respectively,

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1 TUTORIAL 4 Problem 1 An inverted hemispherical bowl of radius R carries a uniform surface charge density σ. Find the potential difference between the north pole and the center. Proof. Computing the potential at the center and pole respectively, V center 1 σ 4πɛ r da 1 σ da 1 σ 4πɛ R 4πɛ R (πr ) σr ɛ and V pole 1 σ 4πɛ r r da with da πr sin θdθ and r R + R R cos θ R (1 cos θ) 1 V pole 1 (σπr ) 4πɛ R t σr πɛ π sin θdθ 1 cos θ σr ( 1 cos θ] π πɛ and so V pole V center σr ɛ ( 1). Problem Two points chargers of the opposite sign Q 1 > and Q < are placed along the z-axis at the positions h 1 and h respectively above a grounded conducting xy-plane, a) Find the magnitude and direction of the force experienced by the charge Q b) Determine the electrostatic energy of the system. 1 Use Law of Cosines with angle the usual angle θ in spherical coordinates 1

2 TUTORIAL 4 Proof. a) The system of images consists of two charges Q 1 and Q placed at (,, h 1 ) and (,, h ) respectively. The total force on charges Q is Q 1 Q F a z 4πɛ(h h 1 ) a Q z 4πɛ(h ) + a Q 1 Q z 4πɛ(h + h 1 ) here we used Coulomb s law and the Superposition principle. b) The energy of the system corresponds to the energy of the system of images with the assumption that the potential below the plane is always zero W E 1 (Q 1V 1 + Q V Q 1 () Q ()). Applying the superposition principle, the potential at the position of Q 1 due to the other charges is V 1 V Q 4πɛ (h 1 h ) Q 1 Q 8πɛ h 1 4πɛ (h 1 + h ) Q 1 4πɛ (h 1 h ) Q Q 1 8πɛ h 4πɛ (h 1 + h ) and so, W E 1 (Q 1V 1 + Q V ) Q 1 Q 4πɛ (h 1 h) Q 1 Q Q 1 Q 16πɛ h 1 16πɛ h 4πɛ (h 1 + h) Problem A point charge Q is located at point (a,, b) between two semi-infinite conducting planes intersecting at right angles. Determine the potential at point P (x, y, z) and the force on Q. Proof. Three image charges are necessary to satisfy the conditions

3 TUTORIAL 4 and so the potential is computing using the Superposition principle: V Q [ ] 4πɛ r 1 r r r 4 where r 1 [(x a) + y + (z b) ] 1 r [(x + a) + y + (z b) ] 1 r [(x + a) + y + (z + b) ] 1 The net force on Q is r 4 [(x a) + y + (z + b) ] 1 F F 1 + F + F Q Q 4πɛ (b) a z 4πɛ (a) a x + Q (a a x + b a z ) 4πɛ ((a) + (b) ) Problem 4 A cylindrical resistor of cross-sectional area A and length L is made from a material with conductivity σ. If the potential is constant over each end and the potential difference between the ends is V what current flows? Proof. As I S J ds JA σea σa L V and so the current density is uniform if the electric field is uniform. To prove this, we know within the cylinder V obeys Laplace s equation with boundary conditions Left End: V L is constant, and we may set V L Right End: V R is constant, and we may set V R V. Cylindrical Surface J n (otherwise the charge leaks out and this is non-conducting).

4 4 TUTORIAL 4 As E n this implies V n with V or its normal derivative specified on all surfaces, the potential is uniquely determined. By guessing one potential that obeys Laplace s equation ( 1 ρ V ) + 1ρ ( ) ρ ρ ρ ρ V φ + ρ V z and the boundary conditions, the unique solution is V V z L where z is measured along the z-axis. The corresponding field is E V V L a z which is uniform, and I is uniform as well. Problem 7 a) Two metal objects are embedded in weakly conducting material of conductivity σ. Show that the resistance between them is related to the capacitance of the arrangement by R ɛ σc b) Suppose a battery is connected between 1 and and charged them up to a potential difference V. If the battery is disconnected, the charge will leak off. Show that V (t) V e t/τ and find the time constant τ in terms of ɛ and σ. Proof. a) I J da where the integral is taken over a surface enclosing the positively charged conductor. Since J σe, Gauss Law implies E da 1 ɛ Q then I σ E da σ Q ɛ but Q CV and V IR so b) Since Q CV CIR then I σ ɛ CIR, or R ɛ σc dq dt I 1 RC Q Q(t) Q e t RC. Or equivalently since V Q/C, V (t) V e t RC τ RC ɛ/σ. and so the time constant is Problem 6 In a vacuum diode, electrons are boiled off a hot cathode, at potential zero, and accelerated across a gap to the anode, which is held at positive potential V. The cloud of moving electrons within the gap (called space charge) quickly builds up to the point where it reduces the field at the surface of the cathode to zero. From then on a steady current I flows between the plates.

5 TUTORIAL 4 5 Suppose the plates are large relative to the separation (A >> d ) so that the edge effects can be neglected. Then V, ρ and v (the speed of the electrons) are all functions of x alone. a) Write Poisson s equation for the region between the plates b) Assuming the electrons start from rest at the cathode, what is their speed at point x, where the potential is V (x)? c) In the steady state, I is independent of x. What, then, is the relation between ρ and v? d) Use these three results to obtain a differential equation for V, by eliminating ρ and v. e) Solve this equation for V as a function of x, V, and d. Plot V (x) and compare it to the potential without space-charge. Find ρ and v as functions of x. f) Show that I KV and find the constant K. This is called the Child-Langmuir law. It holds for other geometries as well, whenever space-charge limits the current. Notice that the space-charge limited diode is nonlinear - it does not obey Ohm s law. Proof. a) b) c) V ρ ɛ d V dx qv 1 mv v 1 ɛ ρ qv m dq dq Aρdx ; dt aρdx dt Aρv I This is a constant, and since ρ is negative, I will be as well. d) d V dx 1 ρ 1 I ɛ ɛ Av I m ɛa qv

6 6 TUTORIAL 4 and so where β I m ɛ A q. e) Multiply by V dv dx, then V dv dx βv 1 dv dx V dv β d V dx βv 1 V 1 dv 1 V βv 1 + Constant. However V () V () (cathode is at potential zero, and field at cathode is zero), so the constant is zero, and V 4βV 1 dx βv 1 4 V 1 4 dv βdx; V 1 4 dv β dx 4 V 4 βx + Constant dv Now V (), so this constant is also zero. ( ) 4 ( ) ( V βx, and so, V (x) β x, or V (x) 4 β 4 81I m x ɛ A q x 4 f) ( In terms of V instead of I this is V (x) V x ( d would increase linearly V (x) V x ) d. V (d) V d V ρ ɛ dx ɛ 1 4 V d 4 1 x 4ɛ V. 9(d x) q qv ( x ) v V m m d Solving for I, this yields ) 4, without space-charge, V ( 8I ) 1 m ɛ d 4 V A q 81md4 ɛ A q I ; I ɛ A q 81md 4 V I 4 ɛ A q 9 V md KV ).