A point charge $+q$ is located at a distance $d$ above an infinite, grounded conducting plane lying in the $xy$-plane. To analyze the electrostatic field in the region above the plane, the grounded conductor is replaced by an equivalent image charge $-q$ positioned at a distance $d$ directly below the plane at position $(0, 0, -d)$. The electric field magnitude just outside the surface of a conductor in electrostatic equilibrium is related to the surface charge density by $E = \dfrac{|\sigma|}{\varepsilon_0}$. Which of the following expressions gives the magnitude of the induced surface charge density $|\sigma(r)|$ on the conducting plane as a function of the radial distance $r$ from the point directly beneath the charge? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Relate the surface charge density to the perpendicular component of the electric field at the surface of the conductor. \[ \|\sigma(r)\| = \varepsilon_0 E_z(r) \]	The question asks for $\|\sigma(r)\|$, and for a conductor in electrostatic equilibrium, the field just outside is entirely perpendicular to the surface with magnitude $E = \dfrac{\|\sigma\|}{\varepsilon_0}$.
Calculate the distance $R$ from both the real charge $+q$ at $(0,0,d)$ and the image charge $-q$ at $(0,0,-d)$ to a point $P$ on the plate at radial distance $r$. \[ R = \sqrt{r^2 + d^2} \]	The electric field magnitude from a point charge depends on the distance squared, $R^2$.
Resolve the electric field vector components at point $P$ due to both charges. \[ E_z = 2 E_1 \sin\theta = 2 \left( \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{R^2} \right) \sin\theta \]	Parallel components along the plate cancel by symmetry, while perpendicular components pointing into the plate (in the $-z$ direction) add constructively.
Substitute the geometric relation $\sin\theta = \dfrac{d}{R} = \dfrac{d}{\sqrt{r^2 + d^2}}$. \[ E_z = \dfrac{2q}{4\pi\varepsilon_0 (r^2 + d^2)} \left( \dfrac{d}{\sqrt{r^2 + d^2}} \right) = \dfrac{qd}{2\pi\varepsilon_0 (r^2 + d^2)^{3/2}} \]	Expressing $\sin\theta$ in terms of given spatial variables converts the geometric field projection into explicit coordinates.
Multiply by $\varepsilon_0$ to evaluate the final surface charge density expression. \[ \|\sigma(r)\| = \varepsilon_0 E_z = \dfrac{qd}{2\pi (r^2 + d^2)^{3/2}} \]	This executes the final link between field strength and induced surface charge.

Why each choice is correct or incorrect:

(A) This choice considers only the perpendicular electric field contribution from the point charge $+q$ and forgets that the image charge $-q$ contributes an equal perpendicular field component, missing a factor of 2.

(B) This is the correct answer.

(C) This choice uses $\cos\theta = \dfrac{r}{\sqrt{r^2+d^2}}$ instead of $\sin\theta = \dfrac{d}{\sqrt{r^2+d^2}}$, effectively computing the tangential component of the field (which cancels to zero) rather than the normal component.

(D) This choice incorrectly squares the distance factor $R^2$ when substituting the geometric angle factor, yielding an incorrect power of $(r^2+d^2)^2$ in the denominator.

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Beginner

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Intermediate

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A long, straight cylindrical wire of radius $R$ carries a steady total current $I$ distributed uniformly across its circular cross section. The permeability of free space is $\mu_0$. Which of the following expressions represents the total magnetic energy stored per unit length inside the volume of the wire?

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Step	Reasoning
Relate the surface charge density to the perpendicular component of the electric field at the surface of the conductor. \[ \|\sigma(r)\| = \varepsilon_0 E_z(r) \]	The question asks for \(\|\sigma(r)\|\), and for a conductor in electrostatic equilibrium, the field just outside is entirely perpendicular to the surface with magnitude \(E = \dfrac{\|\sigma\|}{\varepsilon_0}\).
Calculate the distance \(R\) from both the real charge \(+q\) at \((0,0,d)\) and the image charge \(-q\) at \((0,0,-d)\) to a point \(P\) on the plate at radial distance \(r\). \[ R = \sqrt{r^2 + d^2} \]	The electric field magnitude from a point charge depends on the distance squared, \(R^2\).
Resolve the electric field vector components at point \(P\) due to both charges. \[ E_z = 2 E_1 \sin\theta = 2 \left( \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{R^2} \right) \sin\theta \]	Parallel components along the plate cancel by symmetry, while perpendicular components pointing into the plate (in the \(-z\) direction) add constructively.
Substitute the geometric relation \(\sin\theta = \dfrac{d}{R} = \dfrac{d}{\sqrt{r^2 + d^2}}\). \[ E_z = \dfrac{2q}{4\pi\varepsilon_0 (r^2 + d^2)} \left( \dfrac{d}{\sqrt{r^2 + d^2}} \right) = \dfrac{qd}{2\pi\varepsilon_0 (r^2 + d^2)^{3/2}} \]	Expressing \(\sin\theta\) in terms of given spatial variables converts the geometric field projection into explicit coordinates.
Multiply by \(\varepsilon_0\) to evaluate the final surface charge density expression. \[ \|\sigma(r)\| = \varepsilon_0 E_z = \dfrac{qd}{2\pi (r^2 + d^2)^{3/2}} \]	This executes the final link between field strength and induced surface charge.

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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