A thin ring of radius $R$ lies in the $xy$-plane and is centered at the origin. The ring carries a non-uniform linear charge density given by $\lambda(\theta) = \lambda_0 \cos(2\theta)$, where $\theta$ is the azimuthal angle measured counterclockwise from the positive $x$-axis and $\lambda_0$ is a positive constant. Which of the following expressions correctly gives the electric potential $V(z)$ as a function of position $z$ along the central $z$-axis, relative to $V = 0$ at infinity? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Express the electric potential at point $P = (0,0,z)$ as a line integral over the charge distribution on the ring. \[ V(z) = \dfrac{1}{4\pi\varepsilon_0} \int \dfrac{dq}{r} \]	The question asks for the electric potential $V(z)$ on the central axis due to a continuous non-uniform linear charge density $\lambda(\theta)$.
Relate the distance $r$ from an arbitrary charge element to the field point $P$. \[ r = \sqrt{R^2 + z^2} \]	To evaluate the potential integral, we must determine how the distance $r$ from a charge element on the ring to $(0,0,z)$ depends on position along the ring.
Substitute $dq = \lambda(\theta) R\,d\theta$ and factor out constant terms from the integral. \[ V(z) = \dfrac{1}{4\pi\varepsilon_0 \sqrt{R^2 + z^2}} \int_{0}^{2\pi} \lambda_0 \cos(2\theta) R\,d\theta = \dfrac{\lambda_0 R}{4\pi\varepsilon_0 \sqrt{R^2 + z^2}} \int_{0}^{2\pi} \cos(2\theta)\,d\theta \]	Because every point on the ring is equidistant from any point on the $z$-axis, $r$ does not depend on $\theta$ and factors completely out of the integral.
Evaluate the angular integral over one full revolution. \[ \int_{0}^{2\pi} \cos(2\theta)\,d\theta = \left[ \dfrac{\sin(2\theta)}{2} \right]_{0}^{2\pi} = 0 \implies V(z) = 0 \]	Completing the integration gives the net charge enclosed by the ring and determines the scalar potential.

Why each choice is correct or incorrect:

(A) Incorrectly treats the charge density as uniform with value $\lambda_0$, yielding a net ring charge of $2\pi R \lambda_0$ instead of integrating $\cos(2\theta)$ over its full periodic cycle.

(B) Integrates the absolute magnitude of the charge density across all four quadrants, failing to account for the negative charge regions that cancel the positive charge regions.

(C) This is the correct answer. Since all charge elements on the ring are equidistant from any point on the $z$-axis, the distance $r = \sqrt{R^2 + z^2}$ factors out of the potential integral, making $V(z)$ proportional to the net charge $Q_{\text{net}} = \int_0^{2\pi} \lambda_0 \cos(2\theta) R\,d\theta = 0$.

(D) Mistakenly includes an extra factor of $R/\sqrt{R^2 + z^2}$ inside the potential integral (confusing scalar potential with the directional projection factor of electric field) and evaluates the integral improperly.

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Intermediate

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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

Mathematical

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Step	Reasoning
Express the electric potential at point \(P = (0,0,z)\) as a line integral over the charge distribution on the ring. \[ V(z) = \dfrac{1}{4\pi\varepsilon_0} \int \dfrac{dq}{r} \]	The question asks for the electric potential \(V(z)\) on the central axis due to a continuous non-uniform linear charge density \(\lambda(\theta)\).
Relate the distance \(r\) from an arbitrary charge element to the field point \(P\). \[ r = \sqrt{R^2 + z^2} \]	To evaluate the potential integral, we must determine how the distance \(r\) from a charge element on the ring to \((0,0,z)\) depends on position along the ring.
Substitute \(dq = \lambda(\theta) R\,d\theta\) and factor out constant terms from the integral. \[ V(z) = \dfrac{1}{4\pi\varepsilon_0 \sqrt{R^2 + z^2}} \int_{0}^{2\pi} \lambda_0 \cos(2\theta) R\,d\theta = \dfrac{\lambda_0 R}{4\pi\varepsilon_0 \sqrt{R^2 + z^2}} \int_{0}^{2\pi} \cos(2\theta)\,d\theta \]	Because every point on the ring is equidistant from any point on the \(z\)-axis, \(r\) does not depend on \(\theta\) and factors completely out of the integral.
Evaluate the angular integral over one full revolution. \[ \int_{0}^{2\pi} \cos(2\theta)\,d\theta = \left[ \dfrac{\sin(2\theta)}{2} \right]_{0}^{2\pi} = 0 \implies V(z) = 0 \]	Completing the integration gives the net charge enclosed by the ring and determines the scalar potential.

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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Unit 1 - Vectors and Kinematics

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