Three point charges are fixed at the vertices of an equilateral triangle with side length $a$, as shown in the figure. Two vertices each hold a charge of $+Q$, while the third vertex holds a charge of $-2Q$. What is the magnitude of the net electric field at the center of the triangle, in terms of $Q$, $a$, and Coulomb's constant $k$? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Intermediate

Step	Reasoning
Identify the target quantity and express the superposition principle for the net electric field. \[\vec{E}_{net} = \vec{E}_{+Q,1} + \vec{E}_{+Q,2} + \vec{E}_{-2Q}\]	The net electric field at the center is the vector sum of the electric fields produced by each of the three point charges.
Determine the distance $r$ from any vertex of an equilateral triangle of side length $a$ to its center. \[r = \dfrac{a}{\sqrt{3}} \implies r^2 = \dfrac{a^2}{3}\]	The magnitude of each field contribution depends on the distance $r$ between the charge and point $P$.
Calculate the vector sum of the electric fields due to the two $+Q$ charges. \[E_{+Q,\text{comb}} = 2 E_Q \cos(60^\circ) = 2 \left(\dfrac{3 k Q}{a^2}\right) \left(\dfrac{1}{2}\right) = \dfrac{3 k Q}{a^2}\]	Both $+Q$ charges produce fields of equal magnitude $E_Q = \dfrac{k Q}{r^2} = \dfrac{3 k Q}{a^2}$ pointing away from their respective vertices. By symmetry, their horizontal components cancel, while their vertical components pointing toward the $-2Q$ vertex add together.
Combine the net field of the two $+Q$ charges with the field from the $-2Q$ charge. \[E_{-2Q} = \dfrac{k \|-2Q\|}{r^2} = \dfrac{2 k Q}{\frac{a^2}{3}} = \dfrac{6 k Q}{a^2}\] \[E_{net} = E_{+Q,\text{comb}} + E_{-2Q} = \dfrac{3 k Q}{a^2} + \dfrac{6 k Q}{a^2} = \dfrac{9 k Q}{a^2}\]	The $-2Q$ charge produces a field pointing directly toward itself (in the exact same direction as $E_{+Q,\text{comb}}$). Thus, their magnitudes add directly.

Why each choice is correct or incorrect:

(A) Uses the side length $a$ as the distance to the center, leading to $E_{net} = \dfrac{3 k Q}{a^2}$.

(B) This is the correct answer.

(C) Adds field magnitudes as scalars ($E_A + E_B + E_C = \dfrac{3kQ}{a^2} + \dfrac{3kQ}{a^2} + \dfrac{6kQ}{a^2} = \dfrac{12kQ}{a^2}$) without accounting for vector component cancellation.

(D) Assumes the center distance is $r = \dfrac{a}{3}$, giving $r^2 = \dfrac{a^2}{9}$ and overestimating the field magnitude.

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Step	Reasoning
Identify the target quantity and express the superposition principle for the net electric field. \[\vec{E}_{net} = \vec{E}_{+Q,1} + \vec{E}_{+Q,2} + \vec{E}_{-2Q}\]	The net electric field at the center is the vector sum of the electric fields produced by each of the three point charges.
Determine the distance \(r\) from any vertex of an equilateral triangle of side length \(a\) to its center. \[r = \dfrac{a}{\sqrt{3}} \implies r^2 = \dfrac{a^2}{3}\]	The magnitude of each field contribution depends on the distance \(r\) between the charge and point \(P\).
Calculate the vector sum of the electric fields due to the two \(+Q\) charges. \[E_{+Q,\text{comb}} = 2 E_Q \cos(60^\circ) = 2 \left(\dfrac{3 k Q}{a^2}\right) \left(\dfrac{1}{2}\right) = \dfrac{3 k Q}{a^2}\]	Both \(+Q\) charges produce fields of equal magnitude \(E_Q = \dfrac{k Q}{r^2} = \dfrac{3 k Q}{a^2}\) pointing away from their respective vertices. By symmetry, their horizontal components cancel, while their vertical components pointing toward the \(-2Q\) vertex add together.
Combine the net field of the two \(+Q\) charges with the field from the \(-2Q\) charge. \[E_{-2Q} = \dfrac{k \|-2Q\|}{r^2} = \dfrac{2 k Q}{\frac{a^2}{3}} = \dfrac{6 k Q}{a^2}\] \[E_{net} = E_{+Q,\text{comb}} + E_{-2Q} = \dfrac{3 k Q}{a^2} + \dfrac{6 k Q}{a^2} = \dfrac{9 k Q}{a^2}\]	The \(-2Q\) charge produces a field pointing directly toward itself (in the exact same direction as \(E_{+Q,\text{comb}}\)). Thus, their magnitudes add directly.

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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Intermediate

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