| 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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A conducting loop is fixed in a region with a magnetic field directed perpendicular to the plane of the loop. The induced electromotive force (EMF) \(\mathcal{E}\) in the loop is measured as a function of time \(t\) and is shown in the graph. At time \(t = 0\), the magnetic flux through the loop is zero (\(\Phi_B = 0\)). Which of the following graphs best represents the magnetic flux \(\Phi_B\) through the loop as a function of time \(t\)?

A rigid circular wire loop of radius \(a\) and total resistance \(R\) lies fixed in a plane perpendicular to a spatially uniform magnetic field. The magnitude of the magnetic field changes with time \(t\) according to \(B(t) = B_0 \left(1 + \dfrac{t}{\tau}\right)^3\), where \(B_0\) and \(\tau\) are positive constants. Which of the following expressions represents the total electric charge \(Q\) that flows past a point in the loop between \(t = 0\) and \(t = \tau\)?

A flat rectangular loop of wire with length \(L\), width \(w\), and resistance \(R\) moves in the \(+x\)-direction at constant speed \(v_0\). At time \(t = 0\), the front edge of the loop enters a magnetic field region extending from \(x = 0\) to \(x = 2L\). Within this region, the magnetic field is directed into the page with spatially varying magnitude \(B(x) = B_0 \dfrac{x}{L}\), where \(B_0\) is a positive constant, and \(B = 0\) elsewhere. Defining counterclockwise current as positive, the induced current \(I(t)\) as a function of time increases linearly from \(0\) to \(+I_0\) for \(0 \le t \le T\), remains constant at \(+I_0\) for \(T \le t \le 2T\), and jumps to \(-I_0\) at \(t = 2T\) before rising linearly to \(0\) at \(t = 3T\), where \(T = \dfrac{L}{v_0}\) and \(I_0 = \dfrac{B_0 w v_0}{R}\). Which of the following claims correctly explains the physical origin of a feature in the \(I(t)\) graph?
An ideal circuit consists of an inductor of inductance \(L\) connected in series with a capacitor of capacitance \(C\). At time \(t = 0\), the capacitor carries an initial charge \(Q_0\) and the current in the circuit is zero. Which of the following graphs best represents the charge \(q(t)\) on the capacitor as a function of time \(t\)?

A square loop of wire with side length \(L\) lies in a region containing a non-uniform magnetic field \(\vec{B} = B_0 \left(\dfrac{x}{L}\right) \hat{k}\), where \(B_0\) is a positive constant and \(x \ge 0\). One edge of the loop is fixed along the \(y\)-axis from \(y = 0\) to \(y = L\). The loop is tilted about the \(y\)-axis by an angle \(\theta\) relative to the \(xy\)-plane. What is the ratio of the magnetic flux through the loop when \(\theta = 60^\circ\) to the magnetic flux through the loop when \(\theta = 0^\circ\)?

A horizontal circular ring of radius \(r\), mass \(m\), and electrical resistance \(R\) falls vertically under the influence of gravity through a region with a non-uniform vertical magnetic field. The vertical component of the magnetic field varies linearly with height \(z\) according to \(B_z(z) = B_0 + bz\), where \(B_0\) and \(b\) are positive constants, and \(z\) is measured upward. Air resistance is negligible. Which of the following expressions correctly represents the magnitude of the ring’s terminal velocity \(v_T\)?

A thin, uniform conducting disk of radius \(R\) rotates in the \(xy\)-plane about a fixed vertical axis through its center with a constant angular speed \(\omega\). A non-uniform magnetic field perpendicular to the plane of the disk is given by \(\vec{B}(r) = B_0 \left(\dfrac{r}{R}\right)^2 \hat{k}\), where \(r\) is the radial distance from the axis of rotation and \(B_0\) is a positive constant. Which of the following expressions represents the magnitude of the induced electromotive force (EMF) between the center of the disk and its outer rim?

A long, straight cylindrical wire of radius \(R\) carries a steady total current \(I\) distributed uniformly across its circular cross section. The permeability of free space is \(\mu_0\). Which of the following expressions represents the total magnetic energy stored per unit length inside the volume of the wire?

A circular region of radius \(R\) in the \(xy\)-plane contains a time-dependent, non-uniform magnetic field directed perpendicular to the plane. Inside the region (\(r \le R\)), the magnitude of the magnetic field is given by \(B(r,t) = C t \left(\dfrac{r}{R}\right)^2\), where \(C\) is a positive constant and \(r\) is the radial distance from the central axis. Outside the region (\(r > R\)), the magnetic field is zero. Which of the following expressions gives the magnitude of the induced electric field \(E(r)\) as a function of distance \(r\) from the central axis inside the region (\(r < R\))?

In the circuit shown, an ideal battery with potential difference \(V_0\) is connected to three resistors (\(R_1 = R\), \(R_2 = 2R\), \(R_3 = R\)), an ideal inductor \(L\), and a switch \(S\). The switch \(S\) has been closed for a long time. At time \(t = 0\), switch \(S\) is opened. Which of the following expressions represents the magnitude of the current \(I(t)\) through resistor \(R_2\) as a function of time \(t\) for \(t \ge 0\)?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque 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 Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [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 |
| Variable | SI 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] |
| Variable | Derived 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.
Start with the given measurement: [katex]\text{5 km}[/katex]
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]
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]
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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