Investigate and quantitatively derive and analyse the interaction between charged particles and uniform electric fields, including: electric field between parallel charged plates E = V/d, acceleration of charged particles by the electric field F_net = ma, F = qE, work done on the charge W = qV, W = qEd, K = (1/2)mv^2

Field strength between parallel plates:

$E = \frac{V}{d} = \frac{250}{0.050} = 5.0 \times 10^3$ V/m.

Force on the electron:

$F = qE = 1.60 \times 10^{-19} \times 5.0 \times 10^3 = 8.0 \times 10^{-16}$ N.

Final speed from the work-energy theorem. All the work done by the field becomes kinetic energy because the electron starts at rest:

$qV = \frac{1}{2} m v^2$
$v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 250}{9.11 \times 10^{-31}}}$
$v = \sqrt{8.78 \times 10^{13}} = 9.37 \times 10^6$ m/s.

Markers reward correct unit conversion (cm to m), the sequence $E$ then $F$ then $v$, and the use of $W = qV$ rather than $W = Fd$ (both give the same answer here, but $W = qV$ is the cleaner route).

The work done on a charge $q$ moving through a potential difference $V$ is $W = qV$, independent of the path or the geometry. By the work-energy theorem, $W = \Delta K$, so the kinetic energy gained is $qV$.

You can also see this from $W = Fd = qEd = q(V/d)d = qV$. The plate separation $d$ cancels: a smaller $d$ gives a stronger field, but the particle travels a shorter distance, so the product $Ed = V$ is unchanged.

Markers reward the algebraic cancellation, the connection to the work-energy theorem, and a clear physical statement that $V$ alone determines the energy gain.

Convert the separation to metres and apply $E = \dfrac{V}{d}$.

$d = 4.0\ \text{cm} = 0.040\ \text{m}$

$E = \dfrac{V}{d} = \dfrac{600}{0.040} = 1.5 \times 10^{4}\ \text{V/m}$.

Marks: one for correctly converting $d$ to metres and substituting into $E = V/d$, one for the answer $1.5 \times 10^4\ \text{V/m}$ with the correct unit.

Apply $F = qE$ with the proton's charge $q = e$.

$F = qE = (1.602 \times 10^{-19})(2.5 \times 10^{4}) = 4.0 \times 10^{-15}\ \text{N}$.

Marks: one for the correct formula with values substituted, one for the answer stated to two significant figures with the unit newton.

All the work done by the field becomes kinetic energy since the electron starts from rest: $W = qV = \Delta K$.

$K = eV = (1.602 \times 10^{-19})(120) = 1.92 \times 10^{-17}\ \text{J}$.

Then $K = \tfrac{1}{2}mv^2$, so $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(1.92 \times 10^{-17})}{9.109 \times 10^{-31}}} = \sqrt{4.22 \times 10^{13}} = 6.5 \times 10^{6}\ \text{m/s}$.

Marks: one for $K = eV = 1.92 \times 10^{-17}\ \text{J}$, one for correctly rearranging $K = \tfrac{1}{2}mv^2$ for $v$, one for $v = 6.5 \times 10^{6}\ \text{m/s}$ with the unit.

(a) The graph is a straight line through the origin, so $K$ is directly proportional to $V$ ($K \propto V$), consistent with $K = qV$ for a fixed charge $q$.

(b) Gradient $= \dfrac{\Delta K}{\Delta V} = \dfrac{(6.4 - 1.6) \times 10^{-17}\ \text{J}}{(400 - 100)\ \text{V}} = \dfrac{4.8 \times 10^{-17}}{300} = 1.6 \times 10^{-19}\ \text{C}$.

(c) Since $K = qV$, the gradient $\Delta K / \Delta V$ equals the charge $q$ of the accelerated particle. The calculated gradient $1.6 \times 10^{-19}\ \text{C}$ matches the elementary charge $e = 1.602 \times 10^{-19}\ \text{C}$ to two significant figures, confirming the particle carries a single electronic charge.

Marks: one for identifying direct proportionality ($K \propto V$, line through the origin), one for a correctly computed gradient with the unit coulombs, one for stating the gradient equals the charge $q$, one for comparing the value to $e$ and concluding it is consistent.

Field. $E = \dfrac{V}{d} = \dfrac{300}{0.015} = 2.0 \times 10^{4}\ \text{V/m}$, directed downward (positive plate on top).

Force and acceleration. The electron is negative, so the force is opposite the field, i.e. upward: $F = eE = (1.602 \times 10^{-19})(2.0 \times 10^{4}) = 3.20 \times 10^{-15}\ \text{N}$.

$a = \dfrac{F}{m_e} = \dfrac{3.20 \times 10^{-15}}{9.109 \times 10^{-31}} = 3.51 \times 10^{15}\ \text{m/s}^2$ (upward).

Time in the field. $t = \dfrac{L}{v_x} = \dfrac{0.060}{2.0 \times 10^{7}} = 3.0 \times 10^{-9}\ \text{s}$.

Deflection. $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(3.51 \times 10^{15})(3.0 \times 10^{-9})^2 = 1.6 \times 10^{-2}\ \text{m}$.

The electron deflects $1.6\ \text{cm}$ toward the positive (top) plate. Since it entered midway with a half-gap of $0.75\ \text{cm}$, it strikes the top plate before leaving the field region.

Marks: one for $E = V/d$ with correct field direction, one for identifying the force on the electron as opposite the field (toward the positive plate) with $a = eE/m_e$, one for $t = L/v_x$ and $y = \tfrac{1}{2}at^2$ substituted correctly, one for $y = 1.6 \times 10^{-2}\ \text{m}$ with the physical check that it exceeds the half-gap.

Band-6 plan. (1) Stage 1: use $qV_1 = \tfrac{1}{2}mv_x^2$ to get the entry speed $v_x$. (2) Stage 2: find the transverse acceleration $a = qE/m = qV_2/(md)$ and the transit time $t = L/v_x$. (3) Combine into $y = \tfrac{1}{2}at^2$ and simplify to show the explicit dependence on $V_1$, $V_2$, $d$, $L$. (4) Explain the physical role of each stage (accelerate vs steer) and state how $y$ scales with each variable.

Model answer. In the electron gun, all the work done by the accelerating field becomes kinetic energy: $qV_1 = \tfrac{1}{2}mv_x^2$, so the entry speed into the deflecting plates is $v_x = \sqrt{2qV_1/m}$. This stage sets the horizontal speed only; it does no vertical work.

In the deflecting plates, the field is $E = V_2/d$, giving a transverse force $F = qE = qV_2/d$ and a constant transverse acceleration $a = qV_2/(md)$. The horizontal speed $v_x$ is unchanged (no horizontal force), so the electron crosses the plate length $L$ in time $t = L/v_x$.

The vertical deflection is $y = \tfrac{1}{2}at^2 = \dfrac{1}{2} \cdot \dfrac{qV_2}{md} \cdot \dfrac{L^2}{v_x^2}$. Substituting $v_x^2 = 2qV_1/m$ gives

$y = \dfrac{1}{2} \cdot \dfrac{qV_2}{md} \cdot \dfrac{L^2 m}{2qV_1} = \dfrac{V_2 L^2}{4 d V_1}$.

The charge $q$ and mass $m$ cancel completely, so the deflection depends only on the two voltages and the plate geometry. Physically: $V_1$ sets the transit speed through the deflecting plates (a higher $V_1$ gives a faster electron, less time to be deflected, so $y$ decreases, matching the $1/V_1$ dependence); $V_2$ sets the strength of the deflecting field (a higher $V_2$ gives a larger transverse force, so $y$ increases, matching the direct $V_2$ dependence); $L^2$ reflects that a longer deflecting region gives both a larger acceleration time and, through $y \propto t^2$, a compounding effect; $d$ appears inversely because a wider gap weakens the field for the same $V_2$.

Marker's note: the top band derives the full expression $y = V_2L^2/(4dV_1)$ (not just states $y = \tfrac12 at^2$), explicitly shows $q$ and $m$ cancelling, and interprets the role of each stage (accelerate vs steer) and each variable's effect on $y$, not just the algebra. A response that stops at $y = \tfrac12 at^2$ without the substitution and interpretation caps in the middle band.

Band-6 plan. Structure as a genuine assessment, not a list. (1) Establish the mathematical equivalence (constant transverse acceleration, same SUVAT/parabola derivation). (2) Identify the points of physical difference (nature and controllability of the force, relative magnitude, the effect of switching the field off or reversing it). (3) Reach a balanced judgement on the claim.

Model answer. Mathematically, the claim is well supported. A charge entering a uniform electric field with a constant horizontal velocity $v_x$ experiences a constant transverse acceleration $a = qE/m$, exactly analogous to the constant acceleration $g$ a projectile experiences under gravity. In both cases the horizontal motion is unaffected (constant $v_x$), the vertical motion obeys $y = \tfrac{1}{2}at^2$, and eliminating time gives a parabolic trajectory $y \propto x^2$ in both systems. The equations of motion, and the shape of the path, are identical in form.

However, the two forces differ physically in important ways. Gravity is a fixed, universal, always-attractive force with $g = 9.8\ \text{m/s}^2$ that cannot be switched off or reversed for a given experiment. The electric force $F = qE$ depends on the sign and magnitude of the charge and on the field the experimenter sets up: reversing the plate polarity reverses the direction of deflection instantly, and increasing $V$ increases the acceleration, neither of which is possible with gravity in a school laboratory. In addition, the electric acceleration in a typical CRT or deflection-plate problem ($\sim 10^{14}$-$10^{15}\ \text{m/s}^2$) is many orders of magnitude larger than $g$, which is exactly why gravity is neglected when analysing charged-particle deflection.

This controllability is also the practical payoff: because $E$, and hence $a$, can be set precisely and changed rapidly, electric deflection plates were used to steer electron beams in oscilloscopes and old television tubes, something impossible with a gravitational analogue.

On balance, the claim is justified as a mathematical statement (same equations, same parabolic shape) but overstated as a physical one, since the origin, controllability and typical scale of the force are fundamentally different from gravity.

Marker's note: the top band explicitly separates the mathematical equivalence from the physical differences (rather than treating the whole answer as "they're the same" or "they're different"), quantifies at least one difference (typical acceleration scale or controllability), and ends with an explicit, justified judgement on the claim. A response giving only the $y \propto x^2$ derivation with no assessment of the claim caps in the middle band.