ARCHETYPE: Charged Particle in a Uniform Field

The deflecting-plate geometry never left; it just changed cargo. Continuous inkjet printers charge microscopic ink droplets and fly them between deflection plates — equation (5), a million droplets per second, each steered to its letter-shaped landing spot. Electrostatic precipitators in power-station chimneys charge ash particles and sweep them sideways out of the flue gas — a will-it-reach-the-plate problem run deliberately to "hit." Mass spectrometers and electron microscopes use exactly this kinematics in their injection and steering stages, and the Millikan balance is reborn in labs levitating single nanospheres to hunt for fractional charges — every result so far still lands on the integer ladder. Thomson's sliding glow-spot became the twentieth century's television; the physics you now own steered every pixel of it.

Geometry first. Two parallel plates, oppositely charged, face each other across a gap: between them the field $E$ is uniform — same strength, same direction everywhere (edges excused). A particle of charge $q$ and mass $m$ enters this region. Everything that follows hangs on one observation: the force $F = qE$ is the same at every point of the journey, so the acceleration

where: $q$ — the charge (sign included: positive charges accelerate along $\vec E$, negative ones — electrons — opposite to it); $E$ — the field in $\mathrm{N/C}$ (equivalently $\mathrm{V/m}$); $m$ — the particle's mass. Constant. And "constant acceleration" is a password: it unlocks the entire kinematics toolkit — $v = u + at$, $s = ut + \tfrac12at^2$, $v^2 = u^2 + 2as$ — with no new physics whatsoever. The right way to organise every problem is the dictionary you already trust from Projectile Motion:

The last row is the one conceptual novelty, and examiners guard it jealously. Gravity is democratic: feathers and hammers fall together. Electricity is not: the acceleration $qE/m$ carries the particle's identity. Same field, same entry speed — an electron and a proton bend by wildly different amounts (mass ratio $1836$), and in opposite directions. The field reads each particle's charge-to-mass ratio like a fingerprint; that is exactly how Thomson measured $e/m$ before anyone had seen an electron.

Read the drawing with the sign chain that earns marks: field lines run from the positive plate (top) to the negative plate (bottom), so $\vec E$ points down; the particle drawn is positive, so its force $q\vec E$ also points down, and the path bends toward the negative plate — as shown. Now flip the cargo: send an electron through the same plates and the force reverses, $\vec F = q\vec E$ with $q < 0$, so the electron bends upward, toward the positive plate. That one flip is the archetype's most-harvested mark. Fix the chain in this order, every time: plate signs → field direction ($+\to-$) → force $q\vec E$ with the sign of $q$ → bend. Instinct guesses this at roughly even odds; the chain never misses. One more reading discipline: field lines are not train tracks. The particle does not travel along $\vec E$ — it accelerates along $\pm\vec E$ while keeping whatever sideways velocity it brought, and the path (a parabola) cuts across the field lines the whole way through.

Split the motion into components the moment the problem starts — one axis along the field, one across it.

Across the field (call it $x$): no force component, so uniform velocity. If the particle enters with speed $v_0$ perpendicular to $\vec E$ and the field region (plates) has length $L$, the crossing time is pinned by this free direction alone:

where: $t$ never sees the field, the charge, or the mass. Every deflection question begins here — exactly as every projectile question begins with time-of-flight from one clean direction.

Along the field (call it $y$): uniform acceleration from rest (for perpendicular entry), so the deflection inside the plates is $y_1 = \tfrac12 a t^2$. Substituting (1) and (2):

where: read the scalings like an examiner — deflection grows with the square of plate length, and falls with the square of entry speed. Doubling $v_0$ quarters the bend: fast particles are stiff. The $q/m$ out front is the fingerprint again: at equal speeds, big deflection means light or heavily charged. The particle exits along a straight line, tilted at the exit angle:

where: $\alpha$ — the angle between the exit velocity and the entry direction. If a screen waits a further distance $D$ beyond the plates, the extra deflection is $y_2 = D\tan\alpha$, and the total is

where: the bracket carries a lovely geometric secret: the exiting particle travels as if it had gone straight until the centre of the plates and then turned sharply — the straight-line extension of the exit path passes through the plate midpoint. Thomson used exactly this to convert a screen-spot displacement into $e/m$.

The balance configuration. Turn the setup vertical: field up or down, gravity always down. A charge can then hang motionless when the electric force cancels the weight:

where: this is Millikan's oil-drop condition. Solve for $q = mg/E$: measure the droplet's weight and the balancing field, and you have measured its charge. Millikan's punchline, half a decade in the making: every droplet's charge was an integer multiple of one number, $e = 1.6\times10^{-19}$ C. Charge is grainy.

The energy route. When a question asks for speed after moving a distance $d$ along the force, skip kinematics: the work–energy theorem gives $qEd = \tfrac12mv^2 - \tfrac12mu^2$ in one line. (You will upgrade $qEd$ to "charge times potential difference" in the next chapter — Electrostatic Potential — where this shortcut becomes the definition of the electron-volt.)

When may you ignore gravity?

Every problem above quietly dropped $mg$. Justify it once and carry the licence: an electron in a modest $100\ \mathrm{N/C}$ field feels $qE = 1.6\times10^{-17}$ N against $mg \approx 8.9\times10^{-30}$ N — electricity wins by $\sim10^{12}$. For any subatomic particle in any laboratory field, gravity is noise. The licence expires for oil droplets, dust grains and charged spheres: a $10^{-15}$ kg droplet's weight ($\sim 10^{-14}$ N) is exactly the order of the electric forces used to hold it — the whole point of the Millikan configuration. Rule of thumb: particles — drop gravity; droplets and balls — keep it.

The vertical-field projectile

One more variant closes the family: throw a charged ball at an angle inside a vertical field. Nothing new happens — the two constant accelerations simply add. Field pointing down on a positive charge: effective gravity $g_{\text{eff}} = g + qE/m$; field up: $g_{\text{eff}} = g - qE/m$ (and the ball "falls upward" if $qE > mg$). Every projectile formula then runs with $g_{\text{eff}}$: range $\dfrac{v_0^2\sin2\theta}{g_{\text{eff}}}$, time of flight $\dfrac{2v_0\sin\theta}{g_{\text{eff}}}$. One substitution, entire chapter recycled — the same move as the Atwood machine in a lift.

Board level 1 — proton in a field (CBSE/AP-2 pattern, 3 marks)

A proton ($m = 1.67\times10^{-27}\ \mathrm{kg}$, $q = +1.6\times10^{-19}\ \mathrm{C}$) is released from rest in a uniform field $E = 2.0\times10^{4}\ \mathrm{N/C}$. Find (a) its acceleration, (b) its speed after $1.0\ \mathrm{\mu s}$, and (c) the distance covered in that time. Ignore gravity.

Step 1. Acceleration, equation (1): $a = \dfrac{qE}{m} = \dfrac{(1.6\times10^{-19})(2.0\times10^{4})}{1.67\times10^{-27}} = \boxed{1.9\times10^{12}\ \mathrm{m/s^2}}$, along $\vec E$ (positive charge).

Step 2. Speed from rest: $v = at = (1.92\times10^{12})(1.0\times10^{-6}) = \boxed{1.9\times10^{6}\ \mathrm{m/s}}$.

Step 3. Distance: $s = \tfrac12at^2 = \tfrac12(1.92\times10^{12})(1.0\times10^{-6})^2 = \boxed{0.96\ \mathrm{m}}$. Sanity: the ignore-gravity licence — $qE = 3.2\times10^{-15}$ N versus $mg = 1.6\times10^{-26}$ N — holds by eleven orders of magnitude; and $v \sim 10^6$ m/s is still comfortably non-relativistic, so the kinematics is legal.

Board level 2 — Millikan's balance (ISC/NEET pattern, 3 marks)

An oil droplet of mass $2.94\times10^{-15}\ \mathrm{kg}$ is held stationary between horizontal plates where the field is $6.0\times10^{4}\ \mathrm{N/C}$, directed so as to support the droplet. (a) Find the charge on the droplet. (b) How many elementary charges is that? ($g = 9.8\ \mathrm{m/s^2}$.)

Step 1. Stationary means equilibrium: the electric force balances the weight, equation (6): $qE = mg$.

Step 2. Solve: $q = \dfrac{mg}{E} = \dfrac{(2.94\times10^{-15})(9.8)}{6.0\times10^{4}} = \boxed{4.8\times10^{-19}\ \mathrm{C}}$.

Step 3. Count the packets: $n = \dfrac{q}{e} = \dfrac{4.8\times10^{-19}}{1.6\times10^{-19}} = \boxed{3}$ elementary charges — the droplet is three electrons short (or over, depending on sign). Sanity: the integer landed exactly — as it must. Millikan's actual evidence for quantisation was precisely that every droplet gave a near-integer $n$; an exam answer of $n = 2.7$ means a slipped power of ten, not new physics.

Competitive level 1 — the CRT deflection, plates to screen (JEE-Main pattern, ~3.5 minutes)

An electron enters midway between deflecting plates with horizontal speed $v_0 = 2.0\times10^{7}\ \mathrm{m/s}$. The field between the plates is $E = 3.0\times10^{4}\ \mathrm{N/C}$ and the plates are $L = 4.0\ \mathrm{cm}$ long. (a) Find the deflection as the electron leaves the plates. (b) It then flies $D = 30\ \mathrm{cm}$ to a screen; find the total deflection of the spot. ($m_e = 9.11\times10^{-31}\ \mathrm{kg}$.)

Step 1. Acceleration: $a = \dfrac{eE}{m_e} = \dfrac{(1.6\times10^{-19})(3.0\times10^{4})}{9.11\times10^{-31}} = 5.27\times10^{15}\ \mathrm{m/s^2}$, toward the positive plate.

Step 2. Time inside from the free direction, equation (2): $t = \dfrac{L}{v_0} = \dfrac{0.040}{2.0\times10^{7}} = 2.0\times10^{-9}\ \mathrm{s}$.

Step 3. Deflection inside, equation (3): $y_1 = \tfrac12at^2 = \tfrac12(5.27\times10^{15})(2.0\times10^{-9})^2 = \boxed{1.05\ \mathrm{cm}}$.

Step 4. Exit angle, equation (4): $v_y = at = 1.05\times10^{7}$ m/s, so $\tan\alpha = \dfrac{1.05\times10^{7}}{2.0\times10^{7}} = 0.527$ ($\alpha \approx 28^\circ$).

Step 5. Screen leg plus total, equation (5): $y_2 = D\tan\alpha = (0.30)(0.527) = 15.8\ \mathrm{cm}$, so $y_{\text{total}} = 1.05 + 15.8 = \boxed{16.9\ \mathrm{cm}}$. (Check via the midpoint trick: $(L/2 + D)\tan\alpha = (0.32)(0.527) = 16.9$ cm ✓.) Timing note: the question is four substitutions once you take time from the $x$-direction — resist any urge to chase the $y$-motion for time. Notice how the drift space amplifies: 1 cm of bend inside becomes 17 cm at the screen. That lever arm is why CRTs and oscilloscopes worked with such short plates; budget 3.5 minutes.

Competitive level 2 — will it clear the plates? (JEE-Advanced flavour, ~4 minutes)

An electron enters along the central axis midway between two plates separated by $d = 2.0\ \mathrm{cm}$, of length $L = 10\ \mathrm{cm}$, with speed $v_0 = 3.0\times10^{7}\ \mathrm{m/s}$. Find the maximum field $E$ for which the electron escapes the plates without striking one.

Step 1. Geometry of failure: entering on the axis, the electron may deflect at most $y_{\max} = d/2 = 1.0\ \mathrm{cm}$ before it hits a plate. "Just escapes" means $y_1 = d/2$ exactly at $x = L$.

Step 2. Impose the grazing condition with equation (3): $\dfrac{eE_{\max}L^2}{2m_ev_0^2} = \dfrac{d}{2}$.

Step 3. Solve for the field: $E_{\max} = \dfrac{d\,m_e v_0^2}{e L^2} = \dfrac{(0.020)(9.11\times10^{-31})(3.0\times10^{7})^2}{(1.6\times10^{-19})(0.10)^2}$.

Step 4. Evaluate: numerator $= 0.020 \times 9.11\times10^{-31} \times 9.0\times10^{14} = 1.64\times10^{-17}$; denominator $= 1.6\times10^{-21}$. So $E_{\max} = \boxed{1.0\times10^{4}\ \mathrm{N/C}}$.

Step 5. Check the physics at the limit: with $E = E_{\max}$, $a = 1.8\times10^{15}\ \mathrm{m/s^2}$, $t = L/v_0 = 3.3$ ns, and $y = \tfrac12at^2 = 1.0$ cm ✓ — grazing exit confirmed. Any stronger field and the electron strikes the plate at some $x < L$; note the strike point moves inward as $E$ grows, since $y(x) = \dfrac{eEx^2}{2m_ev_0^2}$ reaches $d/2$ at smaller $x$. Timing note: "does it hit the plate?" always reduces to comparing the equation-(3) deflection with $d/2$ (axis entry) or $d$ (entry grazing one plate) — read the entry position with care, it is where this problem hides its trap. Budget 4 minutes; the algebra is one rearrangement.

Force direction for negative charges. $\vec F = q\vec E$ — the sign of $q$ flips the arrow. Electrons accelerate toward the positive plate, against the field. Every session's easiest lost mark; write the vector equation before touching numbers.

Plate length vs plate separation. $L$ (length along the motion) sets the time and lives inside equation (3); $d$ (gap between plates) only referees whether the particle escapes. Swapping them wrecks will-it-hit problems — as does missing that axis entry allows only $d/2$ of bend.

Hunting time in the wrong direction. Time comes from the force-free direction: $t = L/v_0$, full stop. Solving the quadratic $y = \tfrac12at^2$ for $t$ when $y$ isn't even given is the classic panic spiral. (Same discipline as projectile motion: horizontal clock, vertical bend.)

"All charges fall alike." They do not — the equivalence principle belongs to gravity alone. Acceleration is $qE/m$: an electron bends $\sim1836\times$ more than a proton in the same field, opposite way. MCQs pairing "electron and proton, same field, same deflection" test exactly this.

Dropping gravity for droplets (or keeping it for electrons). The licence is quantitative: compare $qE$ with $mg$. Subatomic particles: gravity loses by $10^{10}$-plus — drop it. Oil drops, dust, pith balls: the forces are engineered comparable — keep it. Stating the comparison in one line is often itself a mark.

Energy shortcut with the wrong distance. $W = qEd$ uses displacement along the field direction only. For perpendicular entry, the across-field travel does zero work; only the deflection $y$ earns kinetic energy. When in doubt, work component-wise.

Rounding the exit angle too early. Carry $\tan\alpha$ unrounded to the screen leg; rounding $0.527$ to $0.5$ shifts worked example 3's spot by 8 mm — outside every tolerance.

Constants. $e = 1.6\times10^{-19}$ C, $m_e = 9.11\times10^{-31}$ kg, $m_p = 1.67\times10^{-27}$ kg are expected memorised for JEE/NEET; CBSE/ISC papers usually restate them. Use $g = 9.8\ \mathrm{m/s^2}$ (or 10 where the paper says so) only in droplet/ball problems — see the licence above.

Texts. NCERT Physics Class 12, Ch. 1 (motion of a charge in a field) · HC Verma, Concepts of Physics Vol. 2, Ch. 29 · AP Physics 2 CED, Electric Force & Field unit · IB Physics guide, topic D.3 (motion of charged particles).

Provenance. All results are standard exam canon; every number recomputed. A verifier should recheck: example 1's $a = 1.92\times10^{12}\ \mathrm{m/s^2}$, $v = 1.9\times10^6$ m/s, $s = 0.96$ m; example 2's $q = 4.8\times10^{-19}$ C $= 3e$; example 3's $t = 2.0$ ns, $y_1 = 1.05$ cm, $\tan\alpha = 0.527$, total $16.9$ cm; example 4's $E_{\max} = 1.0\times10^4$ N/C with grazing check $y = 1.0$ cm.

Prerequisites. The Electric Field · Projectile Motion (mechanics track — the machinery transfers wholesale). Leads to. Force on a Moving Charge (coming in E4) · Crossed Fields and the Velocity Selector (coming in E4) · the electron-volt and energy bookkeeping in Electrostatic Potential (E2).