#import "../template/lib.typ": *
#import "@preview/cetz:0.4.2"
#import "@preview/cetz-plot:0.1.3"
#import "@preview/plotsy-3d:0.2.1": *
#set page(paper: "a4")
#show: notes.with(
  title: "EE1P1",
  subtitle: "Graded Homework Assignment 1",
  author: "Folkert Kevelam"
)

== Exercise 1

=== Part a

The value of the electric field $arrow(E)_cal(p)$ at point $cal(P)(cal(l), cal(l), 0)$
can be calculated by the vector sum of the individual electric fields of $Q_1$, $Q_2$, and
$Q_3$.

The individual vector fields are as follows:

$
  arrow(E)_(1, cal(P)) &= k q / r_1^2 hat(r_1) = k q / (sqrt(cal(l)^2 + cal(l)^2 + 0))^2 1 / (sqrt(cal(l)^2 + cal(l)^2 + 0)) vec(cal(l), cal(l), 0) \
  &= k q / (cal(l)^2) vec(1/(2sqrt(2)), 1/(2sqrt(2)), 0) space [N/C] \
  arrow(E)_(2, cal(P)) &= k (2q) / r_2^2 hat(r_2) = 2 k q / (sqrt(0 + cal(l)^2 + 0))^2 1 / sqrt(0 + cal(l)^2 + 0) vec(0, cal(l), 0) \
  &= k q / cal(l)^2 vec(0,2,0) [N/C]\
  arrow(E)_(2, cal(P)) &= k (-3q) / r_3^2 hat(r_3) = -3 k q / sqrt(cal(l)^2 + 0 + 0)^2 1 / sqrt(cal(l)^2 + 0 + 0) vec(cal(l), 0, 0) \
  &= k q / cal(l)^2 vec(-3,0,0) [N/C]
$

The total electric field at point $cal(P)$ is:

$
  arrow(E)_cal(P) &= arrow(E)_(1, cal(P)) + arrow(E)_(2, cal(P)) + arrow(E)_(3, cal(P)) = k q / cal(l)^2 dot (vec(1/(2sqrt(2)), 1/(2sqrt(2)), 0) + vec(0,2,0) + vec(-3,0,0)) \
  &= k q / cal(l)^2 vec(1/(2sqrt(2)) - 3, 1/(2sqrt(2)) + 2, 0) [N/C]
$

#cetz.canvas({
  import cetz.draw: *
  import cetz-plot: *

  plot.plot(
    size: (10,10),
    axis-style: "school-book",
    x-tick-step: none,
    y-tick-step: none,
    plot-style: (stroke: 2pt),
    {
      plot.add(((0,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
      plot.add(((1,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
      plot.add(((0,1),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: blue))
      plot.add(((1,1),), mark: "o", mark-size: 0.25, mark-style: (stroke: none, fill: black))

      plot.add-anchor("pt", (1,1))

      let vec(data) = {
        plot.add(data, mark: "<>", mark-size: 2)
      }

      let calc_e(charge, distance, v) = {
        let x = (charge / distance) * v.at(0)
        let y = (charge / distance) * v.at(1)

        (x,y)
      }

      let charges = (
        1, 2, -3
      )

      let distances = (
        2, 1, 1
      )

      let dir_vectors = (
        (1/calc.sqrt(2), 1/calc.sqrt(2)),
        (0, 1),
        (1, 0)
      )

      let summed_vector = (0,0)
      let n = 0
      let scale = 0.1

      while n < charges.len() {
        let field = calc_e(charges.at(n), distances.at(n), dir_vectors.at(n))
        let temp_x = field.at(0) + summed_vector.at(0)
        let temp_y = field.at(1) + summed_vector.at(1)

        summed_vector.at(0) = temp_x
        summed_vector.at(1) = temp_y
        n += 1
      }

      vec(((1,1),(1 + scale * summed_vector.at(0), 1 + scale * summed_vector.at(1))))
    }
  )

})

=== Part b

In order to nullify the x-component of the electric field with the charge
$Q_a$, the value of the charge $Q_a$ should be such that:

$
  arrow(E)_(a, cal(P)) hat(x) = - arrow(E)_cal(P) hat(x) \
  k Q_a / r_a^2 hat(r)_a hat(x) = - k q / cal(l)^2 (1/(2sqrt(2)) - 3)
$

For the fourth charge, the vector $r_a$ pointing from charge $Q_a$ to point
$cal(P)$ is equal to:

$
  r_a = vec(3/2 cal(l), 3/2 cal(l), 0) \
  |r_a| = sqrt((3/2 cal(l))^2 + (3/2 cal(l))^2 + 0) = 3/sqrt(2) cal(l) \
  hat(r)_a = vec(1/sqrt(2), 1/sqrt(2))
$

Charge $Q_a$ has the same direction vector as charge $Q_1$, which makes sense
since $Q_1$ lies on the line between $Q_a$ and $cal(P)$. We can plug the vector
parameters into our equality:

$
  k Q_a / (3/sqrt(2) cal(l))^2 vec(1/sqrt(2), 1/sqrt(2), 0) hat(x) &= - k q / cal(l)^2 (1/(2sqrt(2)) - 3) \
  Q_a 2 / 9 1/sqrt(2) &= -q (1/(2sqrt(2)) - 3) \
  Q_a &= -q(9/4 - (27 sqrt(2))/2 ) \
  Q_a &= ((27 sqrt(2)) / 2 - 9/4) q [C]
$

#cetz.canvas({
  import cetz.draw: *
  import cetz-plot: *

  plot.plot(
    size: (10,10),
    axis-style: "school-book",
    x-tick-step: none,
    y-tick-step: none,
    plot-style: (stroke: 2pt),
    {
      plot.add(((0,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
      plot.add(((1,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
      plot.add(((0,1),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: blue))
      plot.add(((1,1),), mark: "o", mark-size: 0.25, mark-style: (stroke: none, fill: black))

      plot.add-anchor("pt", (1,1))

      let vec(data) = {
        plot.add(data, mark: "<>", mark-size: 2)
      }

      let calc_e(charge, distance, v) = {
        let x = (charge / distance) * v.at(0)
        let y = (charge / distance) * v.at(1)

        (x,y)
      }

      let charges = (
        1, 2, -3, (1/4) * (54 * calc.sqrt(2) - 9)
      )

      let distances = (
        2, 1, 1, 9/2
      )

      let dir_vectors = (
        (1/calc.sqrt(2), 1/calc.sqrt(2)),
        (0, 1),
        (1, 0),
        (1/calc.sqrt(2), 1/calc.sqrt(2))
      )

      let summed_vector = (0,0)
      let n = 0
      let scale = 0.1

      while n < charges.len() {
        let field = calc_e(charges.at(n), distances.at(n), dir_vectors.at(n))
        let temp_x = field.at(0) + summed_vector.at(0)
        let temp_y = field.at(1) + summed_vector.at(1)

        summed_vector.at(0) = temp_x
        summed_vector.at(1) = temp_y
        n += 1
      }

      vec(((1,1),(1 + scale * summed_vector.at(0), 1 + scale * summed_vector.at(1))))
    }
  )

})

== Exercise 2

=== Part a

Since the cross-sections of the charge distributions are negligible, we can
assume that the charges can be calculated using a line integral:

Finding charge 1:

$
  Q_1 &= integral_cal(l) lambda_1 (arrow(r)) d arrow(r) \
  Q_1 &= integral_a^(a+L) lambda_1 (r(t)) |r^(')(t)| d t
$

with $r(t) = t dot hat(x) $, $a <= t <= a + L$, $|r^(')(t)| = 1$

$
  Q_1 &= integral_(a)^(a+L) lambda_0 (t-a) / L d t \
  Q_1 &= lambda_0 / L integral_(a)^(a + L) t - a d t \
  Q_1 &= lambda_0 / L [1/2 t^2 - a t]^(a + L)_a \
  Q_1 &= lambda_0 / L ((1/2 (a + L)^2 - a^2 - a L) - (1/2 a^2 - a^2)) \
  Q_1 &= lambda_0 / L ((1/2 a ^2 + a L + 1/2 L^2 - a^2 - a L - 1/2 a^2 + a^2)) \
  Q_1 &= lambda_0 / L (1/2 L^2) = 1/2 lambda_0 L space [C]
$

Finding charge 2:

$
  Q_2 &= integral_cal(l) lambda_2 (arrow(r)) d arrow(r) \
  Q_2 &= integral_a^(a+L) lambda_2 (r(t)) |r^(')(t)| d t
$

with $r(t) = t dot hat(y)$, $a <= t <= a + L$, $|r^(')(t)| = 1$

$
  Q_2 &= integral_(a)^(a+L) Q_0 / t d t \
  Q_2 &= Q_0 integral_(a)^(a+L) 1 / t d t \
  Q_2 &= Q_0 [ln(t)]^(a+L)_a = Q_0 ln((a + L)/a) \
  Q_2 &= Q_0 ln(1 + L/a) space [C]
$

=== Part b



=== Part c

== Exercise 3

#let sigma_0 = 1
#let p_0 = 1
#let a = 0.5
#let xfunc(u,v) = u*calc.cos(v)
#let yfunc(u,v) = u*calc.sin(v)
#let zfunc(u,v) = sigma_0*((u - p_0)/(2*a))*calc.sin(v)

#let scale-factor = 0.25
#let (xscale, yscale, zscale) = (1, 1, 1)
#let scale-dim = (xscale*scale-factor, yscale*scale-factor, zscale*scale-factor)

#plot-3d-parametric-surface(
  xfunc,
  yfunc,
  zfunc,
  xaxis: (0,2),
  yaxis: (0,2),
  zaxis: (0,1),
  subdivisions: 10,
  scale-dim: scale-dim,
  udomain: (p_0, p_0 + 2 * a),
  vdomain: (0, calc.pi / 2),
  axis-step: (1,1,1),
  dot-thickness: 0.05em,
  front-axis-thickness: 0.1em,
  front-axis-dot-scale: (0.04, 0.04),
  rear-axis-dot-scale: (0.08, 0.08),
  axis-label-size: 1.5em,
  xyz-colors: (red, green, blue),
  rotation-matrix: ((-2, 2, 4), (0, -1, 0))
)