A thin, flat, nonconducting annular disk lies in the $xy$-plane centered at the origin. The disk has an outer radius $R$, a central circular hole of radius $\dfrac{R}{2}$, and carries a uniform surface charge density $\sigma$. A point $P$ is located on the positive $z$-axis at a distance $z$ from the origin. Which of the following expressions gives the magnitude of the electric field at point $P$? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Intermediate

Step	Reasoning
Identify the target quantity and divide the continuous charge distribution into element rings. \[ dE_z = \dfrac{1}{4\pi\varepsilon_0} \dfrac{dq \cdot z}{(r^2 + z^2)^{3/2}} \]	The question asks for the electric field along the axis of symmetry of a charged planar shape. A flat annulus can be decomposed into thin concentric rings of radius $r$ and width $dr$.
Express the differential charge element $dq$ in terms of surface charge density $\sigma$. \[ dq = \sigma \cdot dA = \sigma (2\pi r \, dr) \]	To integrate over the surface, $dq$ must be written in terms of the geometry of a ring element of radius $r$ and area $dA = 2\pi r \, dr$.
Substitute $dq$ into the electric field element equation and simplify. \[ dE_z = \dfrac{1}{4\pi\varepsilon_0} \dfrac{\sigma (2\pi r \, dr) z}{(r^2 + z^2)^{3/2}} = \dfrac{\sigma z}{2\varepsilon_0} \dfrac{r \, dr}{(r^2 + z^2)^{3/2}} \]	This provides the differential field contribution $dE_z$ from a single ring element as a function of $r$.
Integrate $dE_z$ over the inner and outer radius boundaries of the annular disk. \[ E_z = \dfrac{\sigma z}{2\varepsilon_0} \int_{R/2}^{R} \dfrac{r \, dr}{(r^2 + z^2)^{3/2}} = \dfrac{\sigma z}{2\varepsilon_0} \left[ -\dfrac{1}{\sqrt{r^2 + z^2}} \right]_{R/2}^{R} \]	The charge distribution extends from the inner hole at $r = \dfrac{R}{2}$ to the outer edge at $r = R$.
Evaluate the definite integral at the limits. \[ E_z = \dfrac{\sigma z}{2\varepsilon_0} \left( \dfrac{1}{\sqrt{\dfrac{1}{4}R^2 + z^2}} – \dfrac{1}{\sqrt{R^2 + z^2}} \right) \]	Evaluating at $r = R$ and $r = \dfrac{R}{2}$ yields the final net electric field expression.

Why each choice is correct or incorrect:

(A) Incorrect because it evaluates the integration limits as (upper limit) minus (lower limit) in reverse order, resulting in a negative magnitude.

(B) Incorrect because it integrates from $r = 0$ to $r = R$, which gives the field of a solid disk and neglects the central circular hole.

(C) Incorrect because it attempts to subtract the area of the hole inside the square root term $\sqrt{R^2 – (R/2)^2 + z^2}$ rather than integrating the ring contributions separately.

(D) This is the correct answer.

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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 $t$?

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Advanced

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Beginner

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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 $t$?

Intermediate

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$A 3D Cartesian coordinate system with x, y, and z axes labeled at the ends of thin black lines with arrowheads. A flat square loop of side length L is situated in the xy-plane with one edge lying along the y-axis from y = 0 to y = L. A second identical square loop shares the same edge along the y-axis but is tilted upward into the xz-plane by an angle \theta relative to the xy-plane. An arc labeled \theta indicates the tilt angle between the two loop planes. Vertical upward arrows parallel to the z-axis represent a magnetic field \vec{B}; the arrows increase in length at larger values of x. No other labels, lines, text, or axes appear.$

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Advanced

Mathematical

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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?

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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 ($r < R$)?

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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 $t \ge 0$?

Step	Reasoning
Identify the target quantity and divide the continuous charge distribution into element rings. \[ dE_z = \dfrac{1}{4\pi\varepsilon_0} \dfrac{dq \cdot z}{(r^2 + z^2)^{3/2}} \]	The question asks for the electric field along the axis of symmetry of a charged planar shape. A flat annulus can be decomposed into thin concentric rings of radius \(r\) and width \(dr\).
Express the differential charge element \(dq\) in terms of surface charge density \(\sigma\). \[ dq = \sigma \cdot dA = \sigma (2\pi r \, dr) \]	To integrate over the surface, \(dq\) must be written in terms of the geometry of a ring element of radius \(r\) and area \(dA = 2\pi r \, dr\).
Substitute \(dq\) into the electric field element equation and simplify. \[ dE_z = \dfrac{1}{4\pi\varepsilon_0} \dfrac{\sigma (2\pi r \, dr) z}{(r^2 + z^2)^{3/2}} = \dfrac{\sigma z}{2\varepsilon_0} \dfrac{r \, dr}{(r^2 + z^2)^{3/2}} \]	This provides the differential field contribution \(dE_z\) from a single ring element as a function of \(r\).
Integrate \(dE_z\) over the inner and outer radius boundaries of the annular disk. \[ E_z = \dfrac{\sigma z}{2\varepsilon_0} \int_{R/2}^{R} \dfrac{r \, dr}{(r^2 + z^2)^{3/2}} = \dfrac{\sigma z}{2\varepsilon_0} \left[ -\dfrac{1}{\sqrt{r^2 + z^2}} \right]_{R/2}^{R} \]	The charge distribution extends from the inner hole at \(r = \dfrac{R}{2}\) to the outer edge at \(r = R\).
Evaluate the definite integral at the limits. \[ E_z = \dfrac{\sigma z}{2\varepsilon_0} \left( \dfrac{1}{\sqrt{\dfrac{1}{4}R^2 + z^2}} – \dfrac{1}{\sqrt{R^2 + z^2}} \right) \]	Evaluating at \(r = R\) and \(r = \dfrac{R}{2}\) yields the final net electric field expression.

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]

Prefix	Symbol	Power of Ten
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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AP Physics

Unit 1 - Vectors and Kinematics

MCQ

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Intermediate

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