275 lines
7.3 KiB
Plaintext
275 lines
7.3 KiB
Plaintext
#import "../template/lib.typ": *
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#import "@preview/cetz:0.4.2"
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#import "@preview/cetz-plot:0.1.3"
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#import "@preview/plotsy-3d:0.2.1": *
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#set page(paper: "a4")
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#show: notes.with(
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title: "EE1P1",
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subtitle: "Graded Homework Assignment 1",
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author: "Folkert Kevelam"
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)
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== Exercise 1
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=== Part a
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The value of the electric field $arrow(E)_cal(p)$ at point $cal(P)(cal(l), cal(l), 0)$
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can be calculated by the vector sum of the individual electric fields of $Q_1$, $Q_2$, and
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$Q_3$.
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The individual vector fields are as follows:
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$
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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) \
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&= k q / (cal(l)^2) vec(1/(2sqrt(2)), 1/(2sqrt(2)), 0) space [N/C] \
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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) \
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&= k q / cal(l)^2 vec(0,2,0) [N/C]\
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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) \
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&= k q / cal(l)^2 vec(-3,0,0) [N/C]
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$
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The total electric field at point $cal(P)$ is:
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$
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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)) \
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&= k q / cal(l)^2 vec(1/(2sqrt(2)) - 3, 1/(2sqrt(2)) + 2, 0) [N/C]
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$
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#cetz.canvas({
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import cetz.draw: *
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import cetz-plot: *
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plot.plot(
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size: (10,10),
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axis-style: "school-book",
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x-tick-step: none,
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y-tick-step: none,
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plot-style: (stroke: 2pt),
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{
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plot.add(((0,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
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plot.add(((1,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
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plot.add(((0,1),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: blue))
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plot.add(((1,1),), mark: "o", mark-size: 0.25, mark-style: (stroke: none, fill: black))
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plot.add-anchor("pt", (1,1))
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let vec(data) = {
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plot.add(data, mark: "<>", mark-size: 2)
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}
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let calc_e(charge, distance, v) = {
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let x = (charge / distance) * v.at(0)
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let y = (charge / distance) * v.at(1)
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(x,y)
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}
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let charges = (
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1, 2, -3
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)
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let distances = (
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2, 1, 1
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)
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let dir_vectors = (
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(1/calc.sqrt(2), 1/calc.sqrt(2)),
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(0, 1),
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(1, 0)
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)
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let summed_vector = (0,0)
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let n = 0
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let scale = 0.1
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while n < charges.len() {
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let field = calc_e(charges.at(n), distances.at(n), dir_vectors.at(n))
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let temp_x = field.at(0) + summed_vector.at(0)
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let temp_y = field.at(1) + summed_vector.at(1)
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summed_vector.at(0) = temp_x
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summed_vector.at(1) = temp_y
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n += 1
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}
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vec(((1,1),(1 + scale * summed_vector.at(0), 1 + scale * summed_vector.at(1))))
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}
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)
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})
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=== Part b
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In order to nullify the x-component of the electric field with the charge
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$Q_a$, the value of the charge $Q_a$ should be such that:
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$
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arrow(E)_(a, cal(P)) hat(x) = - arrow(E)_cal(P) hat(x) \
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k Q_a / r_a^2 hat(r)_a hat(x) = - k q / cal(l)^2 (1/(2sqrt(2)) - 3)
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$
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For the fourth charge, the vector $r_a$ pointing from charge $Q_a$ to point
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$cal(P)$ is equal to:
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$
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r_a = vec(3/2 cal(l), 3/2 cal(l), 0) \
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|r_a| = sqrt((3/2 cal(l))^2 + (3/2 cal(l))^2 + 0) = 3/sqrt(2) cal(l) \
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hat(r)_a = vec(1/sqrt(2), 1/sqrt(2))
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$
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Charge $Q_a$ has the same direction vector as charge $Q_1$, which makes sense
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since $Q_1$ lies on the line between $Q_a$ and $cal(P)$. We can plug the vector
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parameters into our equality:
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$
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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) \
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Q_a 2 / 9 1/sqrt(2) &= -q (1/(2sqrt(2)) - 3) \
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Q_a &= -q(9/4 - (27 sqrt(2))/2 ) \
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Q_a &= ((27 sqrt(2)) / 2 - 9/4) q [C]
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$
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#cetz.canvas({
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import cetz.draw: *
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import cetz-plot: *
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plot.plot(
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size: (10,10),
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axis-style: "school-book",
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x-tick-step: none,
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y-tick-step: none,
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plot-style: (stroke: 2pt),
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{
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plot.add(((0,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
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plot.add(((1,0),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: red))
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plot.add(((0,1),), mark: "o", mark-size: 0.5, mark-style: (stroke: none, fill: blue))
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plot.add(((1,1),), mark: "o", mark-size: 0.25, mark-style: (stroke: none, fill: black))
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plot.add-anchor("pt", (1,1))
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let vec(data) = {
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plot.add(data, mark: "<>", mark-size: 2)
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}
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let calc_e(charge, distance, v) = {
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let x = (charge / distance) * v.at(0)
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let y = (charge / distance) * v.at(1)
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(x,y)
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}
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let charges = (
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1, 2, -3, (1/4) * (54 * calc.sqrt(2) - 9)
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)
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let distances = (
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2, 1, 1, 9/2
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)
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let dir_vectors = (
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(1/calc.sqrt(2), 1/calc.sqrt(2)),
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(0, 1),
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(1, 0),
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(1/calc.sqrt(2), 1/calc.sqrt(2))
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)
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let summed_vector = (0,0)
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let n = 0
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let scale = 0.1
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while n < charges.len() {
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let field = calc_e(charges.at(n), distances.at(n), dir_vectors.at(n))
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let temp_x = field.at(0) + summed_vector.at(0)
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let temp_y = field.at(1) + summed_vector.at(1)
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summed_vector.at(0) = temp_x
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summed_vector.at(1) = temp_y
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n += 1
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}
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vec(((1,1),(1 + scale * summed_vector.at(0), 1 + scale * summed_vector.at(1))))
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}
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)
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})
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== Exercise 2
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=== Part a
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Since the cross-sections of the charge distributions are negligible, we can
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assume that the charges can be calculated using a line integral:
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Finding charge 1:
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$
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Q_1 &= integral_cal(l) lambda_1 (arrow(r)) d arrow(r) \
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Q_1 &= integral_a^(a+L) lambda_1 (r(t)) |r^(')(t)| d t
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$
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with $r(t) = t dot hat(x) $, $a <= t <= a + L$, $|r^(')(t)| = 1$
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$
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Q_1 &= integral_(a)^(a+L) lambda_0 (t-a) / L d t \
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Q_1 &= lambda_0 / L integral_(a)^(a + L) t - a d t \
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Q_1 &= lambda_0 / L [1/2 t^2 - a t]^(a + L)_a \
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Q_1 &= lambda_0 / L ((1/2 (a + L)^2 - a^2 - a L) - (1/2 a^2 - a^2)) \
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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)) \
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Q_1 &= lambda_0 / L (1/2 L^2) = 1/2 lambda_0 L space [C]
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$
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Finding charge 2:
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$
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Q_2 &= integral_cal(l) lambda_2 (arrow(r)) d arrow(r) \
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Q_2 &= integral_a^(a+L) lambda_2 (r(t)) |r^(')(t)| d t
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$
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with $r(t) = t dot hat(y)$, $a <= t <= a + L$, $|r^(')(t)| = 1$
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$
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Q_2 &= integral_(a)^(a+L) Q_0 / t d t \
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Q_2 &= Q_0 integral_(a)^(a+L) 1 / t d t \
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Q_2 &= Q_0 [ln(t)]^(a+L)_a = Q_0 ln((a + L)/a) \
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Q_2 &= Q_0 ln(1 + L/a) space [C]
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$
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=== Part b
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=== Part c
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== Exercise 3
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#let sigma_0 = 1
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#let p_0 = 1
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#let a = 0.5
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#let xfunc(u,v) = u*calc.cos(v)
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#let yfunc(u,v) = u*calc.sin(v)
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#let zfunc(u,v) = sigma_0*((u - p_0)/(2*a))*calc.sin(v)
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#let scale-factor = 0.25
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#let (xscale, yscale, zscale) = (1, 1, 1)
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#let scale-dim = (xscale*scale-factor, yscale*scale-factor, zscale*scale-factor)
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#plot-3d-parametric-surface(
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xfunc,
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yfunc,
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zfunc,
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xaxis: (0,2),
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yaxis: (0,2),
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zaxis: (0,1),
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subdivisions: 10,
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scale-dim: scale-dim,
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udomain: (p_0, p_0 + 2 * a),
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vdomain: (0, calc.pi / 2),
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axis-step: (1,1,1),
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dot-thickness: 0.05em,
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front-axis-thickness: 0.1em,
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front-axis-dot-scale: (0.04, 0.04),
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rear-axis-dot-scale: (0.08, 0.08),
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axis-label-size: 1.5em,
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xyz-colors: (red, green, blue),
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rotation-matrix: ((-2, 2, 4), (0, -1, 0))
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)
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