An electron and a proton are individually launched horizontally with the same initial speed $v_0$ into a uniform electric field $\vec{E}$ between two horizontal parallel plates of length $L$ and separation $d$, as shown. Neither particle strikes either plate while traveling between the plates. How does the kinetic energy gained by the electron, $\Delta K_e$, compare to the kinetic energy gained by the proton, $\Delta K_p$, upon exiting the region between the plates, and why?(a) $\Delta K_e = \Delta K_p$, because both particles carry a charge of magnitude $e$ and traverse a region with the same potential difference $\Delta V$. , (b) $\Delta K_e = \Delta K_p$, because the electron’s smaller mass produces a larger vertical acceleration that exactly cancels the mass dependence in the kinetic energy formula. , (c) $\Delta K_e \Delta K_p$, because both particles experience the same magnitude of electric force for the same time interval, causing the electron to undergo a larger vertical displacement in the direction of the force. :

An electron and a proton are individually launched horizontally with the same initial speed $v_0$ into a uniform electric field $\vec{E}$ between two horizontal parallel plates of length $L$ and separation $d$, as shown. Neither particle strikes either plate while traveling between the plates. How does the kinetic energy gained by the electron, $\Delta K_e$, compare to the kinetic energy gained by the proton, $\Delta K_p$, upon exiting the region between the plates, and why? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Relate the kinetic energy gained by each particle to the work done by the electric field. \[ \Delta K = W = F_E \Delta y \]	The question asks for a comparison of kinetic energy gained ($\Delta K_e$ vs. $\Delta K_p$), which equals the work done by the electrostatic force according to the work-energy theorem.
Determine the time $t$ spent by each particle inside the field region. \[ t = \dfrac{L}{v_0} \]	Since there is no horizontal electric force, the horizontal component of velocity remains constant at $v_0$. The time spent between plates depends only on the plate length $L$ and initial horizontal speed $v_0$.
Calculate the electric force magnitude $F_E$ and vertical acceleration $a$ for a particle of charge magnitude $e$ and mass $m$. \[ F_E = eE \implies a = \dfrac{eE}{m} \]	Both particles carry charge magnitude $e$, so they experience the same force magnitude, but their accelerations differ due to mass.
Calculate the vertical displacement $\Delta y$ and total work done $W$ during time $t$. \[ \Delta y = \dfrac{1}{2} a t^2 = \dfrac{1}{2} \left(\dfrac{eE}{m}\right) \left(\dfrac{L}{v_0}\right)^2 = \dfrac{e E L^2}{2 m v_0^2} \]	Substituting the acceleration and time into kinematic displacement gives the vertical distance shifted, which determines the work done.
Combine work done and electric force to find $\Delta K$ as a function of mass $m$. \[ \Delta K = (eE) \left(\dfrac{e E L^2}{2 m v_0^2}\right) = \dfrac{e^2 E^2 L^2}{2 m v_0^2} \implies \Delta K \propto \dfrac{1}{m} \]	Evaluating $W = F_E \Delta y$ yields the functional dependence of kinetic energy gain on mass.

Why each choice is correct or incorrect:

(A) Incorrect. The particles exit the sides of the plates and do not travel across the full potential difference $\Delta V$ between the plates; their actual potential change depends on their vertical displacement, which is larger for the electron.

(B) Incorrect. While vertical velocity is $v_y = a t \propto 1/m$, kinetic energy depends on $v_y^2 \propto 1/m^2$, so $\Delta K = \frac{1}{2} m v_y^2 \propto m (1/m^2) = 1/m$. Mass does not cancel out.

(C) Incorrect. The electric force depends on charge and field strength ($F_E = eE$), not mass. Both particles experience the exact same electric force magnitude.

(D) This is the correct answer. Both particles spend the same time $t = L/v_0$ in the field and experience the same electric force $F_E = eE$. The smaller mass of the electron leads to a greater vertical acceleration, resulting in a larger vertical displacement $\Delta y$ and therefore more work done by the field ($W = F_E \Delta y$).

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Step	Reasoning
Relate the kinetic energy gained by each particle to the work done by the electric field. \[ \Delta K = W = F_E \Delta y \]	The question asks for a comparison of kinetic energy gained (\(\Delta K_e\) vs. \(\Delta K_p\)), which equals the work done by the electrostatic force according to the work-energy theorem.
Determine the time \(t\) spent by each particle inside the field region. \[ t = \dfrac{L}{v_0} \]	Since there is no horizontal electric force, the horizontal component of velocity remains constant at \(v_0\). The time spent between plates depends only on the plate length \(L\) and initial horizontal speed \(v_0\).
Calculate the electric force magnitude \(F_E\) and vertical acceleration \(a\) for a particle of charge magnitude \(e\) and mass \(m\). \[ F_E = eE \implies a = \dfrac{eE}{m} \]	Both particles carry charge magnitude \(e\), so they experience the same force magnitude, but their accelerations differ due to mass.
Calculate the vertical displacement \(\Delta y\) and total work done \(W\) during time \(t\). \[ \Delta y = \dfrac{1}{2} a t^2 = \dfrac{1}{2} \left(\dfrac{eE}{m}\right) \left(\dfrac{L}{v_0}\right)^2 = \dfrac{e E L^2}{2 m v_0^2} \]	Substituting the acceleration and time into kinematic displacement gives the vertical distance shifted, which determines the work done.
Combine work done and electric force to find \(\Delta K\) as a function of mass \(m\). \[ \Delta K = (eE) \left(\dfrac{e E L^2}{2 m v_0^2}\right) = \dfrac{e^2 E^2 L^2}{2 m v_0^2} \implies \Delta K \propto \dfrac{1}{m} \]	Evaluating \(W = F_E \Delta y\) yields the functional dependence of kinetic energy gain on mass.

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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