An ideal $LC$ circuit consists of an inductor of inductance $L$ and a parallel-plate capacitor of capacitance $C$ with horizontal plates of area $A$. At time $t = 0$, the top plate has charge $+Q_0$, the bottom plate has charge $-Q_0$, and the current in the circuit is zero. The angular frequency of oscillation is $\omega = \dfrac{1}{\sqrt{LC}}$, and the maximum magnitude of the electric field between the capacitor plates is $E_{\text{max}} = \dfrac{Q_0}{A\varepsilon_0}$. At time $t = \dfrac{\pi}{3\omega}$, which of the following correctly identifies the magnitude $E$ of the electric field between the plates and whether the magnitude of the field is increasing or decreasing? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Intermediate

Step	Reasoning
Identify the time-dependent charge function for the capacitor plates in an LC circuit. \[ q(t) = Q_0 \cos(\omega t) \]	The question asks for the electric field magnitude and its rate of change at a specific time, which requires establishing the equation for the charge on the capacitor plates as a function of time.
Relate the charge on the capacitor to the electric field between its plates. \[ E(t) = \dfrac{q(t)}{A\varepsilon_0} = \dfrac{Q_0}{A\varepsilon_0} \cos(\omega t) = E_{\text{max}} \cos(\omega t) \]	The electric field inside a parallel-plate capacitor is proportional to the surface charge density, $E(t) = \dfrac{\sigma(t)}{\varepsilon_0} = \dfrac{q(t)}{A\varepsilon_0}$.
Calculate the electric field magnitude at time $t = \dfrac{\pi}{3\omega}$. \[ E\left(\dfrac{\pi}{3\omega}\right) = E_{\text{max}} \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2} E_{\text{max}} \]	Substituting $t = \dfrac{\pi}{3\omega}$ yields the phase angle $\omega t = \dfrac{\pi}{3}$.
Differentiate $E(t)$ to determine whether the magnitude of the field is increasing or decreasing. \[ \dfrac{dE}{dt} = \dfrac{d}{dt}\left[E_{\text{max}} \cos(\omega t)\right] = -\omega E_{\text{max}} \sin(\omega t) \]	The sign of the derivative $\dfrac{dE}{dt}$ determines whether the magnitude is increasing (positive derivative) or decreasing (negative derivative).
Evaluate the derivative at $t = \dfrac{\pi}{3\omega}$ to conclude the behavior. \[ \left.\dfrac{dE}{dt}\right\|_{t=\frac{\pi}{3\omega}} = -\omega E_{\text{max}} \sin\left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2} \omega E_{\text{max}} < 0 \]	At $\omega t = \dfrac{\pi}{3}$, $\sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} > 0$, making the rate of change negative.

Why each choice is correct or incorrect:

(A) Incorrectly uses $\sin(\pi/3) = \sqrt{3}/2$ to evaluate charge instead of $\cos(\pi/3)$ (confusing current with charge), and assumes the field magnitude is increasing because current magnitude is increasing.

(B) Incorrectly uses $\sin(\pi/3) = \sqrt{3}/2$ to evaluate charge/field magnitude, mistaking the time dependence of the current $I(t) \propto \sin(\omega t)$ for the charge $q(t) \propto \cos(\omega t)$.

(C) This is the correct answer.

(D) Correctly calculates $E = \dfrac{1}{2} E_{\text{max}}$, but incorrectly concludes that the field magnitude is increasing, confusing the growth of current magnitude with the change in electric field.

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Step	Reasoning
Identify the time-dependent charge function for the capacitor plates in an LC circuit. \[ q(t) = Q_0 \cos(\omega t) \]	The question asks for the electric field magnitude and its rate of change at a specific time, which requires establishing the equation for the charge on the capacitor plates as a function of time.
Relate the charge on the capacitor to the electric field between its plates. \[ E(t) = \dfrac{q(t)}{A\varepsilon_0} = \dfrac{Q_0}{A\varepsilon_0} \cos(\omega t) = E_{\text{max}} \cos(\omega t) \]	The electric field inside a parallel-plate capacitor is proportional to the surface charge density, \(E(t) = \dfrac{\sigma(t)}{\varepsilon_0} = \dfrac{q(t)}{A\varepsilon_0}\).
Calculate the electric field magnitude at time \(t = \dfrac{\pi}{3\omega}\). \[ E\left(\dfrac{\pi}{3\omega}\right) = E_{\text{max}} \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2} E_{\text{max}} \]	Substituting \(t = \dfrac{\pi}{3\omega}\) yields the phase angle \(\omega t = \dfrac{\pi}{3}\).
Differentiate \(E(t)\) to determine whether the magnitude of the field is increasing or decreasing. \[ \dfrac{dE}{dt} = \dfrac{d}{dt}\left[E_{\text{max}} \cos(\omega t)\right] = -\omega E_{\text{max}} \sin(\omega t) \]	The sign of the derivative \(\dfrac{dE}{dt}\) determines whether the magnitude is increasing (positive derivative) or decreasing (negative derivative).
Evaluate the derivative at \(t = \dfrac{\pi}{3\omega}\) to conclude the behavior. \[ \left.\dfrac{dE}{dt}\right\|_{t=\frac{\pi}{3\omega}} = -\omega E_{\text{max}} \sin\left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2} \omega E_{\text{max}} < 0 \]	At \(\omega t = \dfrac{\pi}{3}\), \(\sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} > 0\), making the rate of change negative.

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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