AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the target quantity and principle required to solve the problem.
\[ \vec{E}_{net} = \vec{E}_{+} + \vec{E}_{-} \]
The question asks for the electric field \(\vec{E}\) inside a cavity of a non-conducting sphere. Because the cavity disrupts spherical symmetry, direct application of Gauss’s law is difficult; superposition allows us to treat the cavity as a solid sphere of density \(+\rho_0\) overlapping with a smaller sphere of charge density \(-\rho_0\) occupying the cavity volume.
Calculate the electric field \(\vec{E}_+\) at point \(P\) due to a complete solid sphere of charge density \(+\rho_0\) centered at \(O\).
\[ \oint \vec{E}_+ \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \implies E_+ (4\pi r^2) = \frac{\rho_0 \left(\frac{4}{3}\pi r^3\right)}{\varepsilon_0} \implies \vec{E}_+ = \frac{\rho_0 \vec{r}}{3\varepsilon_0} \]
By Gauss’s law, a spherical surface of radius \(r\) encloses charge \(Q_{\text{enc}} = \rho_0 \left(\frac{4}{3}\pi r^3\right)\).
Calculate the electric field \(\vec{E}_-\) at point \(P\) due to a solid sphere of charge density \(-\rho_0\) centered at \(O’\).
\[ \vec{E}_- = \frac{-\rho_0 \vec{r}’}{3\varepsilon_0} = \frac{-\rho_0 (\vec{r} – \vec{a})}{3\varepsilon_0} \]
The position of point \(P\) relative to the cavity center \(O’\) is given by \(\vec{r}’ = \vec{r} – \vec{a}\). Applying Gauss’s law to the sphere of density \(-\rho_0\) yields its field contribution at \(P\).
Superimpose the two field vectors to find the net field \(\vec{E}\) at point \(P\).
\[ \vec{E} = \vec{E}_+ + \vec{E}_- = \frac{\rho_0 \vec{r}}{3\varepsilon_0} – \frac{\rho_0 (\vec{r} – \vec{a})}{3\varepsilon_0} = \frac{\rho_0 \vec{a}}{3\varepsilon_0} \]
Summing the two contributions yields the net field inside the cavity.

Why each choice is correct or incorrect:

(A) This choice considers only the field of the main solid sphere at position \(\vec{r}\), completely neglecting the missing charge within the cavity.

(B) This choice evaluates the field relative to the cavity center \(O’\) using only the negative charge of the cavity region, ignoring the field contributed by the surrounding solid sphere.

(C) This choice adds the field vectors of two positive charge distributions instead of subtracting the field of the removed cavity region.

(D) This is the correct answer. By vector superposition of a positive sphere centered at \(O\) and a negative sphere centered at \(O’\), the position dependence \(\vec{r}\) cancels out, leaving a uniform electric field \(\dfrac{\rho_0 \vec{a}}{3\varepsilon_0}\) throughout the entire cavity.

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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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