A capacitor consists of two rectangular conducting plates of length $L$ and width $W$ tilted at a small angle $\theta$ relative to one another, such that the plate separation increases linearly from $d$ at $x = 0$ to $d + L\theta$ at $x = L$, where $L\theta \ll d$. Let $C_0 = \dfrac{\varepsilon_0 W L}{d}$ be the capacitance of an ideal parallel-plate capacitor of identical dimensions with uniform separation $d$. Which of the following expressions represents the best approximation for the total capacitance $C$ of the tilted-plate capacitor? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify the differential capacitance element $dC$ across the length of the plates. \[ dC = \dfrac{\varepsilon_0 W \, dx}{y(x)} \]	Because the plates are conductors, each infinitesimal cross-section at position $x$ is at a constant potential difference, meaning the infinitesimal strips act as parallel capacitors whose capacitances add directly.
Express the separation distance $y(x)$ as a function of position $x$ and set up the integral for total capacitance $C$. \[ C = \int_0^L \dfrac{\varepsilon_0 W \, dx}{d + x\theta} \]	The gap varies linearly along the length from $d$ at $x = 0$ to $d + L\theta$ at $x = L$ for small angles $\theta$.
Evaluate the integral. \[ C = \dfrac{\varepsilon_0 W}{\theta} \left[ \ln(d + x\theta) \right]_0^L = \dfrac{\varepsilon_0 W}{\theta} \ln\left(\dfrac{d + L\theta}{d}\right) = \dfrac{\varepsilon_0 W}{\theta} \ln\left(1 + \dfrac{L\theta}{d}\right) \]	Integrating $\dfrac{1}{d + x\theta}$ with respect to $x$ yields a logarithmic form.
Apply the Taylor series approximation $\ln(1 + u) \approx u – \dfrac{u^2}{2}$ for $u = \dfrac{L\theta}{d} \ll 1$. \[ C \approx \dfrac{\varepsilon_0 W}{\theta} \left( \dfrac{L\theta}{d} – \dfrac{L^2\theta^2}{2d^2} \right) = \dfrac{\varepsilon_0 W L}{d} \left(1 – \dfrac{L\theta}{2d}\right) = C_0 \left(1 – \dfrac{L\theta}{2d}\right) \]	Since $L\theta \ll d$, expanding to second order in $u$ reveals the leading correction to the standard parallel-plate formula $C_0 = \dfrac{\varepsilon_0 W L}{d}$.

Why each choice is correct or incorrect:

(A) This is the correct answer, obtained by integrating the parallel strip elements along the length and expanding $\ln(1 + u)$ to first order in $\theta$.

(B) Incorrectly predicts an increase in capacitance with tilt, violating the physical principle that increasing the average plate separation decreases capacitance.

(C) Incorrectly evaluates the second term of the Taylor expansion $\ln(1+u) \approx u – \dfrac{u^2}{2}$, omitting the factor of 1/2.

(D) Combines a sign error (predicting capacitance increases) with a missing factor of 1/2 in the series expansion.

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Step	Reasoning
Identify the differential capacitance element \(dC\) across the length of the plates. \[ dC = \dfrac{\varepsilon_0 W \, dx}{y(x)} \]	Because the plates are conductors, each infinitesimal cross-section at position \(x\) is at a constant potential difference, meaning the infinitesimal strips act as parallel capacitors whose capacitances add directly.
Express the separation distance \(y(x)\) as a function of position \(x\) and set up the integral for total capacitance \(C\). \[ C = \int_0^L \dfrac{\varepsilon_0 W \, dx}{d + x\theta} \]	The gap varies linearly along the length from \(d\) at \(x = 0\) to \(d + L\theta\) at \(x = L\) for small angles \(\theta\).
Evaluate the integral. \[ C = \dfrac{\varepsilon_0 W}{\theta} \left[ \ln(d + x\theta) \right]_0^L = \dfrac{\varepsilon_0 W}{\theta} \ln\left(\dfrac{d + L\theta}{d}\right) = \dfrac{\varepsilon_0 W}{\theta} \ln\left(1 + \dfrac{L\theta}{d}\right) \]	Integrating \(\dfrac{1}{d + x\theta}\) with respect to \(x\) yields a logarithmic form.
Apply the Taylor series approximation \(\ln(1 + u) \approx u – \dfrac{u^2}{2}\) for \(u = \dfrac{L\theta}{d} \ll 1\). \[ C \approx \dfrac{\varepsilon_0 W}{\theta} \left( \dfrac{L\theta}{d} – \dfrac{L^2\theta^2}{2d^2} \right) = \dfrac{\varepsilon_0 W L}{d} \left(1 – \dfrac{L\theta}{2d}\right) = C_0 \left(1 – \dfrac{L\theta}{2d}\right) \]	Since \(L\theta \ll d\), expanding to second order in \(u\) reveals the leading correction to the standard parallel-plate formula \(C_0 = \dfrac{\varepsilon_0 W L}{d}\).

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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