A rectangular dielectric slab of dielectric constant $\kappa > 1$ is partially inserted a distance $x$ into a parallel-plate capacitor. The capacitor is connected to a battery that maintains a constant potential difference $V_0$ across the plates, and the slab is free to slide horizontally without friction. Which of the following claims correctly describes the direction of the electrostatic force exerted on the slab by the capacitor and provides the correct justification?(a) The dielectric experiences an outward force, because inserting the slab increases the stored electrostatic energy of the capacitor, and physical systems naturally move toward lower potential energy states. , (b) The dielectric experiences an inward force, because inserting the slab increases the capacitance, and at constant potential difference, the electrostatic force is given by $F = \dfrac{1}{2}V_0^2 \dfrac{dC}{dx} > 0$. , (c) The dielectric experiences an inward force, because inserting the slab decreases the stored electrostatic energy of the capacitor, and the electrostatic force acts in the direction that decreases stored energy. , (d) The dielectric experiences an outward force, because the bound surface charge induced on the dielectric is repelled by the free charge on the capacitor plates. :

A rectangular dielectric slab of dielectric constant $\kappa > 1$ is partially inserted a distance $x$ into a parallel-plate capacitor. The capacitor is connected to a battery that maintains a constant potential difference $V_0$ across the plates, and the slab is free to slide horizontally without friction. Which of the following claims correctly describes the direction of the electrostatic force exerted on the slab by the capacitor and provides the correct justification? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Intermediate

Step	Reasoning
Identify the goal and select the principle relating electrostatic force to energy for a system at constant potential difference.	The question asks for the force on the dielectric slab and its physical justification when the potential difference $V_0$ is held constant by a battery.
Express the stored electrostatic energy of the capacitor as a function of the dielectric insertion distance $x$. \[ U(x) = \dfrac{1}{2} C(x) V_0^2 \]	Because the capacitor is connected to a battery, the potential difference $V_0$ remains constant while the capacitance $C(x)$ varies with insertion depth.
Apply the principle of virtual work at constant potential difference to determine the electrostatic force. \[ F_x = \dfrac{d}{dx}\left(\dfrac{1}{2}C(x)V_0^2\right) = \dfrac{1}{2}V_0^2 \dfrac{dC}{dx} \]	When potential difference is kept fixed by a battery, the battery delivers charge and does work $W_{\text{bat}} = \Delta Q \cdot V_0 = V_0^2 \Delta C$. Taking into account the battery’s work, the net electrostatic force on the movable element is $F_x = +\left(\dfrac{\partial U}{\partial x}\right)_{V_0}$.
Evaluate the sign of $\dfrac{dC}{dx}$ to determine the direction of the force. \[ \dfrac{dC}{dx} > 0 \implies F_x > 0 \]	As the dielectric enters further between the plates, more of the capacitor volume is occupied by a material with dielectric constant $\kappa > 1$, increasing the total capacitance.

Why each choice is correct or incorrect:

(A) Incorrectly identifies the force direction as outward and fails to account for the energy supplied by the battery when determining the direction of electrostatic force at constant potential difference.

(B) This is the correct answer. The force is directed inward because inserting the slab increases capacitance ($dC/dx > 0$), yielding a positive (inward) force $F = \dfrac{1}{2}V_0^2 \dfrac{dC}{dx}$.

(C) Incorrectly claims that stored energy decreases; at constant potential difference $V_0$, stored energy $U = \dfrac{1}{2}CV_0^2$ increases as capacitance increases (confusing constant potential difference with the isolated constant-charge scenario).

(D) Incorrectly states that the force is outward and mischaracterizes the attractive interaction between the free charges on the plates and the opposite bound surface charges induced on the dielectric.

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Step	Reasoning
Identify the goal and select the principle relating electrostatic force to energy for a system at constant potential difference.	The question asks for the force on the dielectric slab and its physical justification when the potential difference \(V_0\) is held constant by a battery.
Express the stored electrostatic energy of the capacitor as a function of the dielectric insertion distance \(x\). \[ U(x) = \dfrac{1}{2} C(x) V_0^2 \]	Because the capacitor is connected to a battery, the potential difference \(V_0\) remains constant while the capacitance \(C(x)\) varies with insertion depth.
Apply the principle of virtual work at constant potential difference to determine the electrostatic force. \[ F_x = \dfrac{d}{dx}\left(\dfrac{1}{2}C(x)V_0^2\right) = \dfrac{1}{2}V_0^2 \dfrac{dC}{dx} \]	When potential difference is kept fixed by a battery, the battery delivers charge and does work \(W_{\text{bat}} = \Delta Q \cdot V_0 = V_0^2 \Delta C\). Taking into account the battery’s work, the net electrostatic force on the movable element is \(F_x = +\left(\dfrac{\partial U}{\partial x}\right)_{V_0}\).
Evaluate the sign of \(\dfrac{dC}{dx}\) to determine the direction of the force. \[ \dfrac{dC}{dx} > 0 \implies F_x > 0 \]	As the dielectric enters further between the plates, more of the capacitor volume is occupied by a material with dielectric constant \(\kappa > 1\), increasing the total capacitance.

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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