Homework #1

Graugnard
MSE 311
28 August 2021

Problem 1

Consider a point charge held a distance d above an infinite grounded conducting plane, as shown below.

(a) Sketch the image charges and calculate the potential in the region above the plate . 
(b) What is the force on the charge q?

Solution:
Mathematically, we need to solve Poisson' s equation in the region of interest (z > 0) . We can assemble a charge configuration that possesses the same boundary conditions as the grounded conducting plane (V = 0 in the xy - plane), if we place an image charge (same magnitude, opposite sign) at an equal distance on the other side of the xy - plane, as shown below .

For this configuration, we can simply write the potential for z > 0 as a superposition of the potential from each point charge .

V[x_,y_,z_,q_,d_]:=(1/(4(Subscript[πϵ, 0])))[(q/(Sqrt[(x^2)+(y^2)+(((z-d))^2)]))+((-q)/(Sqrt[(x^2)+(y^2)+(((z+d))^2)]))];

The force at the charge q can also be written down, but the vector nature must be taken into account .

(Subscript[(OverVector[F]), q])[q_,d_]:=(1/(4(Subscript[πϵ, 0])))(((-(q^2))/(((2d))^2)))(OverHat[z])

Problem 2

Consider an infinitely long rectangular slot consisting of two infinite grounded plates that lie parallel to the xy plane. One plate is located at y = 0 and the other is at y = a. The bounded end of the slot consists of an isolated metal strip held at a potential (Subscript[V, 0])(y). See the drawing below.

(a) Write down the form of Laplace' s equation...

Problem 3

What type of bonding would be expected for each of the following materials: solid xenon, calcium fluoride, bronze, cadmium telluride, rubber, and tungsten.
  
The Pauli exclusion principle states ...

Problem 4

Given that the molar mass of gold is 0.19697 kg mol-1 and its density is 19281 kg m-3, calculate the number of free electrons per unit volume in gold assuming that each atom contributes one electron to the electron gas.

The number of free electrons per unit volume...

Problem 5

Eqn. 2.6 of Tilley is

(Subscript[U, l]) = ((-(Subscript[N, A])((αZ)^2)(e^2))/(4(Subscript[πϵ, 0])r))+...

Using this equation for NaCl, my function is

NaClenergy[r_]:=((-1)/r)+(1/(r^8))

We can plot the energy as a function of atomic separation.

Labeled[Plot[Legended[NaClenergy[r],"Energy"],{r,0,5},AxesLabel->{"Atomic Separation (Å)","Potential Energy (a.u.)"}],Style["Figure 1. Potential energy versus atomic separation for NaCl.","Graphics"]]

The force is the negative derivative of the potential with respect to atomic separation. Thus,

NaClforce[x_]=-D[NaClenergy[x],x] (* note that I used "=" rather than ":=" in order to calculate the derivative now rather than delayed. *)

...