Charge Distribution from a Moving Point Charges: A Naïve Look at Signal Development in Microchannel

Anode Plates

Guilherme Nettesheim, Hervé Grabas

October 30, 2010

1 Introduction

As a first step towards understanding the development of a signal on a transmission line due to a single incident electron, we begin by determining the surface charge and current on an infinite, perfectly conducting plate. The goal is to develop a manageable formula for the electromagnetic field of a constantly accelerating point charge, and an understanding of how charge moves on a conducting surface due to this field. We would then incorporate this understanding into a Monte-Carlo simulation of the signal generated on an anode plate by an electron avalanche from an MCP.

Part I

Electromagnetic Calculations

2 Statics

We begin with the following situation: An electron moving along the z-axis is incident upon a thick, perfectly conducting, grounded, half-infinite metal surface. We assume the electron moves slowly enough that we do not need to consider the finite speed of electromagnetic interaction. PICTURE OF SETUP HERE

2.1 Electric field of a point charge moving constant velocity

If we use cylindrical coordinaters, our boundary condition is that Φ(ρ, θ, z = 0) = 0. This condition is easily satisfied using the using the method of images. If our electron has position r⃗(t) = z_{∘}(t)ẑ at time t and charge q, we place a similar particle at −r⃗(t) with charge −q, giving the potential:

Φ(ρ, θ, z) = \frac{q}{4πϵ_{∘}} (\frac{1}{\sqrt{ρ^{2} + (z − z_{∘}(t))^{2}}} − \frac{1}{\sqrt{ρ^{2} + (z + z_{∘}(t))^{2}}} )

As this potential satisfies our boundary conditions, it is the correct potential. The surface charge σ(ρ, θ) is given by:

σ(ρ, θ) = −ϵ_{∘} \frac{∂Φ}{∂z}|_{z=0} = \frac{−qz_{∘}(t)}{2π(ρ^{2} + z_{∘}(t)^{2})^{3/2}}

2.1.1 Surface current from charge movement

We can consider the current passing through a closed path with area A on the plane’s surface as a function of time by defining the current as:

I(t) = \frac{∂Q_{in}}{∂t}

Where

Q_{in} = ∫_{A}σ dA

If we consider a ring of radius ρ_{∘} centered on the origin, we see that the current through it is:

I(t) = 2π \frac{∂}{∂t} ∫_{ρ_{∘}}σ(ρ, θ)ρdρ = \frac{∂}{∂t} (\frac{qz_{∘}(t)}{\sqrt{ρ^{2} + z_{∘}(t)^{2}}} ) = qż_{∘}(t)((ρ^{2}+z_{∘}(t)^{2})^{−1/2}−z_{∘}(t)^{2}(ρ^{2}+z_{∘}(t)^{2})^{−3/2})

Assuming that the particle is always moving, we see that we would only get zero current if:

z_{∘}(t)^{2} = z_{∘}(t)^{2} + ρ^{2}
z_{∘}(t) = 0

Therefore, so long as the electron is above the plate, positive charge will move inward.

2.2 Magnetic field from a delta-function current: How not to solve the problem

From the above we see that there is indeed a surface current, cause by the changing electric field of the electron. Let us, therefore, consider the electron as the following current:

I_{e}(t, z) = −qδ(z − z_{∘}(t))ẑ

We can therefore solve for potential:

A⃗(ρ, z, t) = \frac{μ_{∘}}{4π} ∫ \frac{I⃗_{e}(t)}{\sqrt{ρ^{2} + (z − z′)^{2}}} dz′ = \frac{−qμ_{∘}}{4π} ẑ∫ \frac{δ(z_{∘}(t) − z′)}{\sqrt{ρ^{2} + (z − z′)^{2}}} dz′ = \frac{−qμ_{∘}}{4π} \frac{ẑ}{\sqrt{ρ^{2} + (z − z_{∘}(t))^{2}}}

and the magnetic field:

B⃗(ρ, z, t) = ∇ × A⃗ = \frac{−qμ_{∘}}{4π} \frac{θ̂}{[ρ^{2} + (z − z_{∘}(t))^{2}]^{3/2}}

This magnetic field has no ẑ component, and therefore is completely parallel to the plane. This gives us zero surface current on the plane, conflicting with what we’ve found above. This is precisely because our current is not steady, and the above potential formulation of the Biot-Savart Law holds only for magnetostatics.

2.3 Insulated plane

The question might now arise as to what would happen if we did not ground our metal surface. We must notice that if we set the potential at infinity as ground and let our plane extend to infinity, that our plane will be grounded as well. To deal with an insulated plane, we must, therefore, make it finite. However, we cannot easily solve the boundary conditions for finite conducting plane with the image charge method. So we solve Laplace’s equation in the space above the plane through Legendre polynomials, with the boundary condition that every surface of the metal slab be at ground. Having solved this problem, we imagine disconnecting our slab from ground. Since we know that the electrostatic forces are now balanced on the metal slab, if we add:

Q_{add} = − ∫_{A}σ dA = q

to the slab, the charge will arrange itself as if on a completely isolated and neutral conductor, with no knowledge of the incident electron or the induced charge on the metal plate. It follows that this additional charge density will be constant in time, and will only depend on the charge of the incident electron and the geometry of the metal slab. In modeling the transmission line, we will have to consider both insulated and grounded cases: Though the transmission line is grounded, charge will not flow from ground instantaneously. We can therefore think of the ends of the transmition line as effectively grounded for all instants, and a segment of the transmission line, extending from the center, as insulated, with this segment shrinking with respect to time. That is, we may think of the entire strip as insulated att = 0 and grounded as t → inf

3 Liénard-Wiechert Fields for a moving point charge

For particles in motion r⃗(t), the Liénard-Wiechert Potentials are more appropriate methods of deriving the electromagnetic field. They are:

Φ(x⃗, t) = \frac{q}{4πϵ_{∘}} [\frac{1}{κR}]_{ret}

A⃗(x⃗, t) = \frac{q}{4πϵ_{∘}} [\frac{β⃗}{κR}]_{ret}

where β⃗ = c^{−1} ˙r⃗(t) , R⃗ = x⃗ − r⃗(t), κ = 1 − R̂ · β⃗, and the subscript ret denotes that the quantity should be evaluated at the retarded time t′ = t − [R(t′)/c]

3.1 Electric field of a point charge moving at constant velocity

We see that the electric field is:

E⃗(x⃗, t) = \frac{q}{4πϵ_{∘}} [\frac{(R̂ − β⃗)(1 − β^{2})}{κ^{3}R^{2}}]_{ret} + \frac{q}{4πϵ_{∘}c} [\frac{R̂}{κ^{3}R} × \{(R̂ − β⃗) × ˙ β⃗\}]_{ret}

As a first approximation, we consider an electron traveling perpendicular to the plane at a constant velocity v. This eliminates the second term, which is linear in ˙ β⃗ and which we denote the acceleration field. The result is:

E⃗(x⃗, t) = \frac{q}{4πϵ_{∘}} [\frac{(R̂ − β⃗)(1 − β^{2})}{κ^{3}R^{2}}]_{ret}

Evaluating for retarded time we see that:

E⃗(x⃗, t) = \frac{q}{4πϵ_{∘}} \frac{(1 − β^{2})}{(1 − β^{2}sin^{2}θ)^{3/2}} \frac{R̂}{R^{2}}

where θ is the angle between R⃗ and v⃗. This electric field, like in the electrostatic case, points radially out. Adding an image charge will, therefore, solve the boundary condition E_{∥} = 0. This will again give us a time-dependent charge distribution on the surface.

3.2 Magnetic field of a point charge moving at constant velocity

The magnetic field is related to the electric field by the following equation:

B⃗ = \frac{1}{c}([R̂]_{ret} × E⃗)

Since

[R̂]_{ret} = \frac{x⃗ − v⃗t′}{[R]_{ret}} = \frac{(x⃗ − v⃗t) + (t − t′)v⃗}{[R]_{ret}} = \frac{R⃗}{[R]_{ret}} + \frac{v⃗}{c}

we have the magnetic field:

B⃗ = \frac{1}{c} (\frac{R⃗}{[R]_{ret}} + \frac{v⃗}{c} ) × E⃗ = \frac{1}{c^{2}} (v⃗ × E⃗)

It is obvious that B⃗ must be everywhere parallel to the plane, since v⃗ is perpendicular to the plane. Thus, we again automatically satisfy the magnetic boundary condition B_{⊥} = 0. To find the induced surface current K⃗, we apply

B⃗|_{surface} = μ_{∘}(K⃗ × n̂)

where n̂ is the unit vector normal to the plane surface, and B⃗ is the magnetic field of the incoming point charge and its image charge. Since K⃗ ⊥ n̂ we have:

K⃗ = n̂ × B⃗ = \frac{1}{c^{2}} [n̂ × (v⃗ × E⃗)] = \frac{1}{c^{2}} [v⃗(n̂ · E⃗) − E⃗(n̂ · v⃗)] = \frac{β}{c} E⃗

Where we used the fact that E⃗ ⊥

Part II

Simulation

4 A first-order model

We present a simple simulation of the signal induced on a single transmission line. We do not doubt that the simplifications made in this simulation will lead to significant difficulties with its final results, but we design it so we may easily incorporate increasingly sophisticated approximations.

4.1 What we are simulating

We simulate the signal detected at a point x = L on the transmission line due to an electron incident normal to its midpoint, x = 0. The electron’s trajectory x(t) ≡ x is due solely to its initial velocity leaving the MCP, v_{∘}, and the acceleration due to the gap voltage differential, a.

4.2 How we simulate it

We take the situation to be two dimensional, with the transmission line consisting of two infinite, conducting lines, and we turn these into a series of discrete capacitors of length l. We think of the approaching electron as charging each capacitor to a voltage V (x).