AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the asymptotic behavior of each ring at distances much larger than their radius. Because \(z \gg R\), higher-order multipole moments (quadrupole, etc.) decay faster than \(z^{-3}\). To leading order in \(1/z\), each ring can be replaced by an effective point charge located at its center: \(+Q\) at \((0,0,a)\) and \(-Q\) at \((b,0,-a)\).
Formulate the electric potential \(V(x,y,z)\) near the \(z\)-axis.
\[V(x,y,z) \approx \dfrac{Q}{4\pi\varepsilon_0} \left( \dfrac{1}{\sqrt{x^2 + y^2 + (z-a)^2}} – \dfrac{1}{\sqrt{(x-b)^2 + y^2 + (z+a)^2}} \right)\]
To calculate the field components \(\vec{E} = -\nabla V\) at \((0,0,z)\), we set up the potential due to the two effective point charges near the observation region.
Evaluate \(V(0,0,z)\) and take the derivative with respect to \(z\) to find \(E_z\).
\[V(0,0,z) \approx \dfrac{Q}{4\pi\varepsilon_0} \left( \dfrac{1}{z-a} – \dfrac{1}{\sqrt{b^2 + (z+a)^2}} \right) \approx \dfrac{Q}{4\pi\varepsilon_0} \left( \left(\dfrac{1}{z} + \dfrac{a}{z^2}\right) – \left(\dfrac{1}{z} – \dfrac{a}{z^2}\right) \right) = \dfrac{2Qa}{4\pi\varepsilon_0 z^2}\]
\[E_z = -\dfrac{d}{dz}\left( \dfrac{2Qa}{4\pi\varepsilon_0 z^2} \right) = \dfrac{4Qa}{4\pi\varepsilon_0 z^3}\]
The \(z\)-component of the electric field is given by \(E_z = -\dfrac{\partial V}{\partial z}\).
Differentiate \(V(x,y,z)\) with respect to \(x\) at \((0,0,z)\) to find \(E_x\).
\[E_x = -\left. \dfrac{\partial V}{\partial x} \right|_{(0,0,z)} = -\dfrac{Q}{4\pi\varepsilon_0} \left( 0 – \dfrac{b}{(b^2 + (z+a)^2)^{3/2}} \right) \approx \dfrac{Qb}{4\pi\varepsilon_0 z^3}\]
The transverse offset \(b\) breaks symmetry along the \(x\)-axis, inducing an \(x\)-component of the field.
Combine components into the final electric field vector.
\[\vec{E} \approx \dfrac{Q}{4\pi\varepsilon_0 z^3} \left( b\,\hat{i} + 4a\,\hat{k} \right)\]
By symmetry in \(y\), \(E_y = 0\). Combining \(E_x\) and \(E_z\) gives the vector expression.

Why each choice is correct or incorrect:

(A) This choice incorrectly assumes the x-component points in the -x direction and fails to multiply by 2 when differentiating 1/z^2 with respect to z.

(B) This choice incorrectly evaluates E_z as 2Qa/(4\pi\varepsilon_0 z^3), missing the factor of 2 that arises from evaluating -d/dz(1/z^2) = 2/z^3.

(C) This is the correct answer.

(D) This choice erroneously doubles the x-component of the field, incorrectly applying the axial factor of 2 to both components.

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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