Difficulty: Introductory review | Prerequisites: P211 kinematics, Newton's second law
This topic reviews how a particle moving in one direction responds to a constant force applied perpendicular to its velocity. It is the mechanical analogue of what happens to a charged particle entering a uniform electric field between parallel plates, which is one of the first setups you will meet in P212. If you are comfortable with projectile motion under gravity, the physics here is identical, just rotated.
A particle enters a region of uniform force at constant horizontal speed and picks up a parabolic vertical displacement, exactly like a ball thrown sideways in a gravitational field. The exit deflection and the maximum force the particle can tolerate before hitting a plate both follow from one kinematic equation. Dimensional analysis and limiting-behaviour checks let you narrow the answer before doing any algebra.
Dimensional analysis
The practice of checking whether both sides of an equation carry the same physical dimensions (length, mass, time, etc.). If they do not, the equation is wrong, full stop. In simple terms, this is a quick sanity check: if one side is in metres and the other in seconds, something has gone off the rails.
Limiting behaviour
Testing an expression by pushing one variable to an extreme (zero, infinity) and asking whether the result makes physical sense. Think of it as: "If I make the plates infinitely long, should the deflection be zero? No? Then the formula is suspect."
Uniform force
A force that has the same magnitude and direction at every point in a region. In simple terms, it is a force that does not change no matter where the particle happens to be, just like gravity near Earth's surface.
Deflection angle (theta)
The angle between the particle's velocity vector and the original horizontal direction at the moment it exits the force region. Think of it as the direction the particle is now heading, measured from where it was heading before it entered the plates.
A particle of mass m travels horizontally at speed v0.
At x = 0 it enters a pair of plates: spacing w, length L.
Between the plates a constant force F acts in the +y direction.
After exiting, the particle moves freely (no force).
This is the exact geometry you will see in P212 for a charged particle in a uniform electric field.
Before computing anything, three candidate expressions were tested:
Candidate (1): y = F / m(v0 + L) fails because v0 (m/s) and L (m) have different dimensions and cannot be added.
Candidate (2): y = FL / mv0 fails because FL has dimensions of energy and mv0 has dimensions of momentum, giving a ratio with dimensions of velocity, not length.
Candidate (3): y = Fw^2 / mv0^2 has correct dimensions but fails the limiting-behaviour test: as L approaches 0, the particle spends no time between the plates, so y should go to zero. This expression does not.
A quick fix for (3): replace w with L. The resulting expression FL^2 / mv0^2 increases with F and L, decreases with m and v0, and is independent of w. All physically sensible. Dimensional analysis alone gets the answer to within a factor of 2.
Horizontal motion: constant velocity v0, so the transit time through the plates is t = L / v0.
Vertical motion: constant acceleration a = F / m, starting from rest in the y direction.
Standard kinematic result: y = (1/2) a t^2.
Substituting: y = (1/2)(F/m)(L/v0)^2 = FL^2 / (2mv0^2).
This confirms the dimensional-analysis estimate and supplies the missing factor of 1/2.
The particle enters on the midplane, so the furthest it can travel vertically is w/2 before hitting a plate.
Setting y = w/2 and solving for F gives Fmax = mv0^2 w / L^2.
At exit, the horizontal velocity is still v0 (no horizontal force).
The vertical velocity picked up is vy = (F/m)(L/v0).
The deflection angle satisfies tan(theta) = vy / vx = FL / (mv0^2).
Under maximum-force conditions (F = Fmax), substituting gives tan(theta) = w / L, which is dimensionless as expected.
Quantity | Formula | Notes |
|---|---|---|
Vertical displacement at exit | y = FL^2 / (2mv0^2) | From y = (1/2)at^2 with a = F/m, t = L/v0 |
Maximum force | Fmax = mv0^2 w / L^2 | Set y = w/2 and solve for F |
Deflection angle | tan(theta) = FL / (mv0^2) | General case |
Deflection angle at Fmax | tan(theta) = w / L | Substituting Fmax into the general expression |
Students often think the displacement depends on plate spacing w. It does not. The spacing only matters for whether the particle hits a plate.
Students sometimes add quantities with different dimensions (e.g. v0 + L) when guessing formulas. Dimensional analysis catches this immediately.
A common error is forgetting that the vertical displacement depends on L^2, not L. The particle accelerates, so doubling the plate length quadruples the deflection.
Students sometimes assume the horizontal velocity changes between the plates. It does not: there is no horizontal component of the force.
This is precisely how a cathode-ray tube (CRT) steers an electron beam. Parallel deflection plates create a uniform electric field, the electron enters horizontally, and the maths above tells you where the spot lands on the screen. The same geometry appears in mass spectrometers and ink-jet printers.
Dimensional analysis is a recurring exam skill in P212. You will be expected to check your answers this way.
The relationship y = FL^2/(2mv0^2) reappears directly in P212 with F = qE (charge times electric field strength).
Always test limiting behaviour: does the result go to zero when it should? Does it blow up when that makes sense physically?
The problem explicitly warns: "Never plug in numbers until the end of your calculation." Symbolic answers are required for checking limits.
True or false: the vertical displacement of the particle depends on the plate spacing w. (False)
Fill in the blank: the transit time through the plates is t = ___ / ___. (L / v0)
True or false: doubling the plate length doubles the vertical deflection. (False, it quadruples it because y is proportional to L^2.)
True or false: the horizontal velocity changes while the particle is between the plates. (False)
Fill in the blank: at maximum force, tan(theta) = ___ / ___. (w / L)
Q: A particle of mass m enters parallel plates of length L at speed v0. A constant force F acts in the +y direction. Derive the y-coordinate at exit.
A: Horizontal transit time is t = L/v0. Vertical acceleration is a = F/m. Using y = (1/2)at^2 gives y = FL^2/(2mv0^2).
Q: For the setup above, what is the maximum value of F such that the particle does not strike either plate (spacing w)?
A: Set y = w/2 and solve: Fmax = mv0^2 w / L^2.
Q: Under maximum-force conditions, what is tan(theta), the tangent of the exit deflection angle?
A: tan(theta) = vy/v0, where vy = (F/m)(L/v0). Substituting Fmax gives tan(theta) = w/L.
Q: A student proposes y = FL/(mv0) as the answer for the exit displacement. Use two physics-based arguments to explain why this cannot be correct.
A: First, FL has dimensions of energy and mv0 has dimensions of momentum, so the ratio has dimensions of velocity, not length. Second, the expression is linear in L, but since the particle accelerates, the displacement should grow faster than linearly with plate length.
Q: Why does the correct expression for y not depend on the plate spacing w?
A: The spacing w describes the geometry of the plates but does not affect the force, the acceleration, or the time spent in the field. It only determines whether the particle strikes a plate.
This connects to P212 charged-particle deflection because replacing F with qE (charge times electric field) turns every formula here into the electrostatics version. It also connects to P211 projectile motion, the horizontal component is constant, the perpendicular component accelerates from rest. The dimensional-analysis and limiting-behaviour techniques recur throughout P212 as primary answer-checking tools.
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