A solid insulating sphere of radius $R$ contains a total net charge $Q$. The volume charge density $\rho$ inside the sphere varies directly with radial distance $r$ from the center, such that $\rho(r) = C r$ for $r \le R$, where $C$ is a constant. In terms of $Q$, $R$, $r$, and fundamental constants, which of the following expressions represents the magnitude of the electric field at a distance $r < R$ from the center of the sphere? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Intermediate

Step	Reasoning
Apply Gauss’s law to a concentric spherical Gaussian surface of radius $r < R$. \[ \oint \vec{E} \cdot d\vec{A} = E(r)(4\pi r^2) = \dfrac{q_{\text{enc}}(r)}{\varepsilon_0} \]	The question asks for the electric field inside a spherically symmetric charge distribution.
Calculate the enclosed charge $q_{\text{enc}}(r)$ by integrating the non-uniform volume charge density. \[ q_{\text{enc}}(r) = \int_0^r \rho(r’) \, dV = \int_0^r (C r’)(4\pi r’^2 \, dr’) = 4\pi C \int_0^r r’^3 \, dr’ = \pi C r^4 \]	To evaluate Gauss’s law, we need the enclosed charge inside radius $r$.
Relate the constant $C$ to the total charge $Q$ of the sphere. \[ Q = q_{\text{enc}}(R) = \pi C R^4 \implies C = \dfrac{Q}{\pi R^4} \]	The final expression must be in terms of given variables $Q$ and $R$ rather than the constant $C$.
Substitute $C$ back into the enclosed charge expression and solve for $E(r)$. \[ q_{\text{enc}}(r) = \left(\dfrac{Q}{\pi R^4}\right) \pi r^4 = Q \dfrac{r^4}{R^4} \]	Combining the enclosed charge with Gauss’s law isolates the electric field magnitude $E(r)$.
Solve Gauss’s law for $E(r)$. \[ E(r) \cdot 4\pi r^2 = \dfrac{1}{\varepsilon_0} \left( Q \dfrac{r^4}{R^4} \right) \implies E(r) = \dfrac{Q r^2}{4\pi \varepsilon_0 R^4} \]	Dividing enclosed charge by $4\pi \varepsilon_0 r^2$ yields the final expression.

Why each choice is correct or incorrect:

(A) This is the correct answer, derived by integrating $\rho(r) = Cr$ over a spherical volume element to determine enclosed charge as a function of total charge $Q$.

(B) Incorrectly assumes electric field is proportional to $r$, which holds for uniform charge density but fails for a non-uniform distribution.

(C) Incorrectly determines $C$ by setting $C R = Q / (\frac{4}{3}\pi R^3)$ using average density rather than evaluating $Q = \int \rho \, dV$.

(D) Results from an integration error in powers of $r$ when calculating total charge versus enclosed charge.

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Step	Reasoning
Apply Gauss’s law to a concentric spherical Gaussian surface of radius \(r < R\). \[ \oint \vec{E} \cdot d\vec{A} = E(r)(4\pi r^2) = \dfrac{q_{\text{enc}}(r)}{\varepsilon_0} \]	The question asks for the electric field inside a spherically symmetric charge distribution.
Calculate the enclosed charge \(q_{\text{enc}}(r)\) by integrating the non-uniform volume charge density. \[ q_{\text{enc}}(r) = \int_0^r \rho(r’) \, dV = \int_0^r (C r’)(4\pi r’^2 \, dr’) = 4\pi C \int_0^r r’^3 \, dr’ = \pi C r^4 \]	To evaluate Gauss’s law, we need the enclosed charge inside radius \(r\).
Relate the constant \(C\) to the total charge \(Q\) of the sphere. \[ Q = q_{\text{enc}}(R) = \pi C R^4 \implies C = \dfrac{Q}{\pi R^4} \]	The final expression must be in terms of given variables \(Q\) and \(R\) rather than the constant \(C\).
Substitute \(C\) back into the enclosed charge expression and solve for \(E(r)\). \[ q_{\text{enc}}(r) = \left(\dfrac{Q}{\pi R^4}\right) \pi r^4 = Q \dfrac{r^4}{R^4} \]	Combining the enclosed charge with Gauss’s law isolates the electric field magnitude \(E(r)\).
Solve Gauss’s law for \(E(r)\). \[ E(r) \cdot 4\pi r^2 = \dfrac{1}{\varepsilon_0} \left( Q \dfrac{r^4}{R^4} \right) \implies E(r) = \dfrac{Q r^2}{4\pi \varepsilon_0 R^4} \]	Dividing enclosed charge by \(4\pi \varepsilon_0 r^2\) yields the final 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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