| Step | Reasoning |
|---|---|
| Identify the target quantity and express potential as the integral of the electric field from reference to point r. \[ V(r) = V(R) – \int_R^r E_{in}(r’) \, dr’ \] |
To find V(r) inside the sphere relative to V(\infty) = 0, we must integrate the electric field from infinity to r in two segments: outside (r > R) and inside (r \le R). |
| Use Gauss’s law to find the total charge Q and the electric field outside the sphere (r >= R), then compute V(R). \[ Q_{tot} = \int_0^R \rho(r’) 4\pi r’^2 \, dr’ = \int_0^R (Cr’) (4\pi r’^2) \, dr’ = 4\pi C \int_0^R r’^3 \, dr’ = \pi C R^4 \] \[ E_{out}(r) = \dfrac{Q_{tot}}{4\pi \varepsilon_0 r^2} = \dfrac{\pi C R^4}{4\pi \varepsilon_0 r^2} = \dfrac{C R^4}{4\varepsilon_0 r^2} \] \[ V(R) = -\int_{\infty}^R E_{out}(r’) \, dr’ = -\int_{\infty}^R \dfrac{C R^4}{4\varepsilon_0 r’^2} \, dr’ = \dfrac{C R^3}{4\varepsilon_0} \] |
The boundary value V(R) is required to determine the integration constant at the sphere’s surface. |
| Use Gauss’s law to find the electric field inside the sphere (r <= R). \[ Q_{enc}(r) = \int_0^r (Cr’) (4\pi r’^2) \, dr’ = \pi C r^4 \] \[ E_{in}(r) \cdot 4\pi r^2 = \dfrac{Q_{enc}(r)}{\varepsilon_0} = \dfrac{\pi C r^4}{\varepsilon_0} \implies E_{in}(r) = \dfrac{C r^2}{4\varepsilon_0} \] |
The integrand E_{in}(r) is determined by the charge enclosed within a radius r. |
| Integrate E_{in}(r) from R to r and add V(R) to obtain the final potential V(r). \[ V(r) – V(R) = -\int_R^r \dfrac{C r’^2}{4\varepsilon_0} \, dr’ = -\dfrac{C}{4\varepsilon_0} \left[ \dfrac{r’^3}{3} \right]_R^r = \dfrac{C}{12\varepsilon_0} (R^3 – r^3) \] \[ V(r) = \dfrac{C R^3}{4\varepsilon_0} + \dfrac{C}{12\varepsilon_0} (R^3 – r^3) = \dfrac{C}{12\varepsilon_0} (4R^3 – r^3) \] |
This combines the surface potential contribution with the potential change inside the sphere. |
Why each choice is correct or incorrect:
(A) Calculates \(V(r) – V(R)\) correctly as \(\dfrac{C}{12\varepsilon_0}(R^3 – r^3)\) but forgets to add the non-zero surface potential \(V(R) = \dfrac{CR^3}{4\varepsilon_0}\).
(B) Neglects the surface potential \(V(R)\) and fails to divide by 3 when integrating \(r^2\,dr\).
(C) This is the correct answer.
(D) Fails to use the 3D volume element \(4\pi r^2 dr\) when evaluating the enclosed charge, leading to an incorrect polynomial coefficient for \(r^3\).
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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 \(t \ge 0\)?
C
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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] | |
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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