A cylindrical capacitor of length $L$ consists of a solid conducting inner cylinder of radius $a$ and a thin conducting outer cylindrical shell of radius $c$. The space between the conductors is filled with two coaxial cylindrical layers of dielectric materials: the inner layer from radius $a$ to radius $b$ has dielectric constant $\kappa_1$, and the outer layer from radius $b$ to radius $c$ has dielectric constant $\kappa_2$. Assuming edge effects are negligible ($L \gg c$), which of the following expressions represents the capacitance of this capacitor? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Relate capacitance to free charge $Q$ and potential difference $\Delta V$, noting that calculating $\Delta V$ requires integrating the electric field across both dielectric layers. \[C = \dfrac{Q}{\Delta V} = \dfrac{Q}{\int_a^c E(r)\,dr} = \dfrac{Q}{\int_a^b E_1(r)\,dr + \int_b^c E_2(r)\,dr}\]	To find capacitance, we must express the potential difference $\Delta V = V(a) – V(c)$ as a function of the enclosed free charge $Q$.
Apply Gauss’s law in dielectrics to find the radial electric field magnitude in each region given free charge $Q$ on the inner cylinder. \[\oint \kappa \vec{E} \cdot d\vec{A} = \dfrac{Q_{\text{free}}}{\varepsilon_0} \implies \kappa E(r) (2\pi r L) = \dfrac{Q}{\varepsilon_0}\] \[E_1(r) = \dfrac{Q}{2\pi \varepsilon_0 \kappa_1 L r} \quad (a < r < b)\] \[E_2(r) = \dfrac{Q}{2\pi \varepsilon_0 \kappa_2 L r} \quad (b < r < c)\]	The presence of a dielectric reduces the electric field magnitude by a factor of $\kappa$ for a given free charge.
Integrate the electric field piecewise from $r = a$ to $r = c$ to calculate the total potential difference $\Delta V$. \[\Delta V = \int_a^b \dfrac{Q}{2\pi \varepsilon_0 \kappa_1 L r}\,dr + \int_b^c \dfrac{Q}{2\pi \varepsilon_0 \kappa_2 L r}\,dr\] \[\Delta V = \dfrac{Q}{2\pi \varepsilon_0 L} \left[ \dfrac{1}{\kappa_1} \ln\left(\dfrac{b}{a}\right) + \dfrac{1}{\kappa_2} \ln\left(\dfrac{c}{b}\right) \right]\]	The total potential difference across the capacitor is the sum of the potential drops across each dielectric layer.
Substitute $\Delta V$ back into the capacitance formula and simplify. \[C = \dfrac{Q}{\dfrac{Q}{2\pi \varepsilon_0 L} \left[ \dfrac{1}{\kappa_1} \ln\left(\dfrac{b}{a}\right) + \dfrac{1}{\kappa_2} \ln\left(\dfrac{c}{b}\right) \right]} = \dfrac{2\pi \varepsilon_0 L}{\dfrac{1}{\kappa_1} \ln\left(\dfrac{b}{a}\right) + \dfrac{1}{\kappa_2} \ln\left(\dfrac{c}{b}\right)}\]	Dividing $Q$ by $\Delta V$ cancels the charge $Q$ and yields the geometric capacitance expression.

Why each choice is correct or incorrect:

(A) This expression incorrectly assumes the dielectrics are in parallel across the same potential difference and simply averages/adds their dielectric constants over the total radius ratio.

(B) This expression incorrectly uses $E = \kappa E_0$ instead of $E = E_0 / \kappa$, putting dielectric constants in the numerator of the field and potential integral rather than the denominator.

(C) This expression incorrectly multiplies the log factors in the denominator, failing to execute the sum of potential drops across series dielectric regions.

(D) This is the correct answer derived from Gauss’s law in dielectrics and integrating the electric field across both concentric regions.

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Step	Reasoning
Relate capacitance to free charge \(Q\) and potential difference \(\Delta V\), noting that calculating \(\Delta V\) requires integrating the electric field across both dielectric layers. \[C = \dfrac{Q}{\Delta V} = \dfrac{Q}{\int_a^c E(r)\,dr} = \dfrac{Q}{\int_a^b E_1(r)\,dr + \int_b^c E_2(r)\,dr}\]	To find capacitance, we must express the potential difference \(\Delta V = V(a) – V(c)\) as a function of the enclosed free charge \(Q\).
Apply Gauss’s law in dielectrics to find the radial electric field magnitude in each region given free charge \(Q\) on the inner cylinder. \[\oint \kappa \vec{E} \cdot d\vec{A} = \dfrac{Q_{\text{free}}}{\varepsilon_0} \implies \kappa E(r) (2\pi r L) = \dfrac{Q}{\varepsilon_0}\] \[E_1(r) = \dfrac{Q}{2\pi \varepsilon_0 \kappa_1 L r} \quad (a < r < b)\] \[E_2(r) = \dfrac{Q}{2\pi \varepsilon_0 \kappa_2 L r} \quad (b < r < c)\]	The presence of a dielectric reduces the electric field magnitude by a factor of \(\kappa\) for a given free charge.
Integrate the electric field piecewise from \(r = a\) to \(r = c\) to calculate the total potential difference \(\Delta V\). \[\Delta V = \int_a^b \dfrac{Q}{2\pi \varepsilon_0 \kappa_1 L r}\,dr + \int_b^c \dfrac{Q}{2\pi \varepsilon_0 \kappa_2 L r}\,dr\] \[\Delta V = \dfrac{Q}{2\pi \varepsilon_0 L} \left[ \dfrac{1}{\kappa_1} \ln\left(\dfrac{b}{a}\right) + \dfrac{1}{\kappa_2} \ln\left(\dfrac{c}{b}\right) \right]\]	The total potential difference across the capacitor is the sum of the potential drops across each dielectric layer.
Substitute \(\Delta V\) back into the capacitance formula and simplify. \[C = \dfrac{Q}{\dfrac{Q}{2\pi \varepsilon_0 L} \left[ \dfrac{1}{\kappa_1} \ln\left(\dfrac{b}{a}\right) + \dfrac{1}{\kappa_2} \ln\left(\dfrac{c}{b}\right) \right]} = \dfrac{2\pi \varepsilon_0 L}{\dfrac{1}{\kappa_1} \ln\left(\dfrac{b}{a}\right) + \dfrac{1}{\kappa_2} \ln\left(\dfrac{c}{b}\right)}\]	Dividing \(Q\) by \(\Delta V\) cancels the charge \(Q\) and yields the geometric capacitance expression.

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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Advanced

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