An ideal LC circuit consists of an inductor of inductance $L$ and a parallel-plate capacitor filled completely with a dielectric slab of dielectric constant $\kappa > 1$. The circuit oscillates with an initial angular frequency $\omega_0$ and a maximum capacitor charge $Q_0$. At the exact instant when the current in the circuit reaches its maximum value, the dielectric slab is instantaneously pulled out of the capacitor without altering the charge on the plates at that instant. What are the new angular frequency $\omega'$ of oscillation and the new maximum charge $Q_{\max}'$ on the capacitor during subsequent oscillations? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify the expression for the initial angular frequency $\omega_0$ in terms of the dielectric constant $\kappa$ and empty capacitance $C_0$. \[ \omega_0 = \dfrac{1}{\sqrt{L C_i}} = \dfrac{1}{\sqrt{L \kappa C_0}} \]	The question asks for the new angular frequency $\omega’$ after the dielectric is removed, so we must first relate $\omega_0$ to the initial capacitance $C_i = \kappa C_0$.
Determine the new angular frequency $\omega’$ after the dielectric slab is removed. \[ \omega’ = \dfrac{1}{\sqrt{L C_0}} = \sqrt{\kappa} \cdot \dfrac{1}{\sqrt{L \kappa C_0}} = \sqrt{\kappa}\omega_0 \]	Removing the dielectric reduces the capacitance to $C_f = C_0$.
Analyze the total energy stored in the circuit at the exact instant the dielectric is removed. \[ \text{At } q = 0, \quad U_{\text{total}} = U_L = \dfrac{1}{2} L I_{\max}^2 = U_0 = \dfrac{Q_0^2}{2 \kappa C_0} \]	To find the new maximum charge $Q_{\max}’$, we must evaluate the system’s total stored energy at the moment of dielectric removal when $q = 0$.
Equate the conserved magnetic energy to the new maximum electric energy in the empty capacitor to solve for $Q_{\max}’$. \[ \dfrac{(Q_{\max}’)^2}{2 C_0} = \dfrac{Q_0^2}{2 \kappa C_0} \implies (Q_{\max}’)^2 = \dfrac{Q_0^2}{\kappa} \implies Q_{\max}’ = \dfrac{Q_0}{\sqrt{\kappa}} \]	Because $q = 0$ at the moment of removal, the electric field between the plates is zero, so no electrostatic force acts on the dielectric and no work is done on it during removal ($W_{\text{ext}} = 0$). Total energy $U_0$ is conserved and subsequently transfers entirely into the empty capacitor $C_0$.

Why each choice is correct or incorrect:

(A) Assumes angular frequency scales as $1/\kappa$ instead of $1/\sqrt{C}$, and incorrectly treats maximum charge as independent of the capacitance change.

(B) Correctly calculates $\omega’$, but treats maximum charge as conserved despite the capacitance dropping while all energy was stored in the magnetic field.

(C) Incorrectly treats removing the dielectric as an increase in capacitance, reversing the frequency factor to $\omega_0/\sqrt{\kappa}$ and predicting an increased maximum charge.

(D) This is the correct answer. Removing the dielectric reduces capacitance to $C_0$, increasing frequency by a factor of $\sqrt{\kappa}$. Since $q=0$ at removal, no work is done, conserving total energy $U = \dfrac{Q_0^2}{2\kappa C_0}$, which yields $Q_{\max}’ = \dfrac{Q_0}{\sqrt{\kappa}}$.

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Step	Reasoning
Identify the expression for the initial angular frequency \(\omega_0\) in terms of the dielectric constant \(\kappa\) and empty capacitance \(C_0\). \[ \omega_0 = \dfrac{1}{\sqrt{L C_i}} = \dfrac{1}{\sqrt{L \kappa C_0}} \]	The question asks for the new angular frequency \(\omega’\) after the dielectric is removed, so we must first relate \(\omega_0\) to the initial capacitance \(C_i = \kappa C_0\).
Determine the new angular frequency \(\omega’\) after the dielectric slab is removed. \[ \omega’ = \dfrac{1}{\sqrt{L C_0}} = \sqrt{\kappa} \cdot \dfrac{1}{\sqrt{L \kappa C_0}} = \sqrt{\kappa}\omega_0 \]	Removing the dielectric reduces the capacitance to \(C_f = C_0\).
Analyze the total energy stored in the circuit at the exact instant the dielectric is removed. \[ \text{At } q = 0, \quad U_{\text{total}} = U_L = \dfrac{1}{2} L I_{\max}^2 = U_0 = \dfrac{Q_0^2}{2 \kappa C_0} \]	To find the new maximum charge \(Q_{\max}’\), we must evaluate the system’s total stored energy at the moment of dielectric removal when \(q = 0\).
Equate the conserved magnetic energy to the new maximum electric energy in the empty capacitor to solve for \(Q_{\max}’\). \[ \dfrac{(Q_{\max}’)^2}{2 C_0} = \dfrac{Q_0^2}{2 \kappa C_0} \implies (Q_{\max}’)^2 = \dfrac{Q_0^2}{\kappa} \implies Q_{\max}’ = \dfrac{Q_0}{\sqrt{\kappa}} \]	Because \(q = 0\) at the moment of removal, the electric field between the plates is zero, so no electrostatic force acts on the dielectric and no work is done on it during removal (\(W_{\text{ext}} = 0\)). Total energy \(U_0\) is conserved and subsequently transfers entirely into the empty capacitor \(C_0\).

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

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