Today in Physics 17: boundary conditions and electrostatic boundary-value problems Boundary conditions in electrostatics Simple solution of Poisson s equation as a boundary-value problem: the space-charge limited vacuum diode Vx ( i ) ρ ( x i ) 0.1 0.05 0 0 0 0 0.05 0.1 x i 40 0 0.05 0.1 0.015 x i vx ( i ) c 0.01 0.005 0 0 0.05 0.1 x i 7 September 00 Physics 17, Fall 00 1
Electrostatic boundary-value problems We have encountered several differential equations relating the field, scalar potential, and charge density, that we will in general want to solve for E or V. But the solutions aren t unique unless enough constraints boundary values are supplied. If the potential, or components of the field, are specified on some surface or line (e.g. equipotentials), a unique solution of the differential equation(s) for V or E in the rest of the volume or area will be obtained. This will usually be how solutions of Laplace s equation will proceed. Specification of the charge density results in specification of boundary conditions on field and potential, as follows. (This often comes in when one solves the Poisson equation.) 7 September 00 Physics 17, Fall 00
Surface charges and electric-field boundary conditions Consider a surface with charge per unit area σ, not necessarily constant. What is the relation of this charge to the electric field on either side of the surface? E above Zoom in on an area a ε small enough to consider σ flat, and draw a E below Gaussian surface with planar symmetry that has sides ε ( 0) perpendicular to the surface. Then, since the flux through the sides is negligible, E d a = 4π Qenclosed E E a= 4πσ a E E = 4πσ ( ) ( ), above, below, above, below The charge sheet makes a discontinuity of 4 πσ in E. 7 September 00 Physics 17, Fall 00 3
Surface charges and electric-field boundary conditions (continued) ( ) Now consider a loop, ε 0 by L, perpendicular to ε the surface, bisected by the surface, and small σ L enough that the surface is flat over its dimensions. In a line integral of field along this loop the contribution of the short sides is negligibly small, so E d l = 0 ( ) E E L = 0 E E = 0, above, below, above, below The parallel component of E is continuous across the sheet of charge. E above E below 7 September 00 Physics 17, Fall 00 4
Surface charges and electric-potential boundary conditions Implications of this on V: For two points a and b, ±ε from the surface, V V = E dl ε above below 0 0 i.e. V is continuous across the charged sheet. Consider the gradient of V. ( ) ( ) Or, with n as the coordinate axis perpendicular to the surface at this point, above below ˆ V V V V n = 4πσ = 4πσ n n 7 September 00 Physics 17, Fall 00 5 b a V V = E E = πσ above below, above, below 4 ( ) above below
Griffiths problem!.48: 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 0. The cloud of moving electrons within the gap (called the 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. Suppose the plates are large relative to the separation d, so that edge effects can be neglected. Then V, ρ, and v (the speed of the electrons) are functions of x alone. (a) Solve Poisson s equation for the potential between the anode and cathode, as a function of x, V 0, and d. 3 (b) Show that I = KV 0, and find the constant K. This expression is called Child s Law. 7 September 00 Physics 17, Fall 00 6
(continued) Heater Cathode e - V = 0 V = V 0 v d A Anode x As soon as electrons boil off in large numbers, their presence slows down further electron ejection because they re so slow at first. (This is what is meant by space-charge limited.) The space charge is denser near the cathode. Although the electrons move, the space charge distribution is constant after a while, so this is an electrostatics problem. 7 September 00 Physics 17, Fall 00 7
(continued) Poisson s equation, in one dimension: ( x) Boundary conditions: Solution: V 0 = 0, V d = V0, E 0 = 0 First, consider a charge q that makes it to where the potential is V. The field has done work W = qv, so 1 W = qv = ( KE) = mv v = qv m Also consider a slice of the space charge, with thickness : dq = ρ A dq = A Av dt = dt dv V = = 4πρ ( ) ( ) ( ) 7 September 00 Physics 17, Fall 00 8
(continued) But dq/dt = I, which is constant if the charge distribution has reached a steady state, so dv I I m 1 = 4πρ = 4π = 4π = βv Av A qv I m where β = 4 π. A q To solve, substitute dv = V, and multiply through by V : dv 1 V = βv V dv 1dV V = βv 7 September 00 Physics 17, Fall 00 9
(continued) Integrate with respect to x: dv 1dV V = β V 1 1 V = βv + C dv But V(0) = 0, and E( 0) = 0= ( 0) = V ( 0 ), so C = 0, and dv 14 V = = βv 14 V dv = β 4 34 V = β x+ D 3 7 September 00 Physics 17, Fall 00 10
(continued) V(0) = 0 once again mandates that D = 0. Also V d 4 34 3 V0 = βd β = V0 3 3d ( ) = V 0, so 43 43 43 3 3 34 x V( x) = β x = V0 x = V0 3d d That s the solution of part (a). To work out the current (part b), start by getting the charge density and speed: 1 dv 1 V0 4 1 3 V0 ρ = 4 = 4 d x = π π 4 3 3 3 3 9π d x qv qv0 x v = = m m d 3 ( ) 7 September 00 Physics 17, Fall 00 11
(continued) Thus I VA 0 qv0 x = ρ Av = 3 m d 9π ( d x) 3 A q = V = 9π d m 3 3 0 KV0, where K = A q. 9π d m Child s Law Note that the current and voltage are not directly proportional, as in Ohm s law. 7 September 00 Physics 17, Fall 00 1