Figure shows a narrow beam of electrons entering with a velocity of $3 \times 10^7$ m/s, symmetrically through the space between two parallel horizontal plates $P_1 P_1'$ and $P_2 P_2'$ kept 2 cm apart. If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end $P_2'$ .

Q: Figure shows a narrow beam of electrons entering with a velocity of $3 \times 10^7$ m/s, symmetrically through the space between two parallel horizontal plates $P_1 P_1'$ and $P_2 P_2'$ kept 2 cm apart. If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end $P_2'$.

**Data & setup** - Entry speed (horizontal) $v = 3 \times 10^7$ m/s; plate length $l = 3\text{ cm} = 0.03\text{ m}$; separation $d = 2\text{ cm} = 0.02\text{ m}$. - Entering midway, the beam must deflect $y = d/2 = 0.01$ m to just hit end $P_2'$. - $m = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C. **Motion (like a projectile)** - Time between plates: $t = l/v = 0.03 / (3 \times 10^7) = 1 \times 10^{-9}$ s. - Vertical acceleration: $a = eE/m = eV/(m d)$. - Deflection: $y = \frac{1}{2} a t^2 = \frac{1}{2} \cdot eV/(md) \cdot t^2 = d/2$. **Solve for V** $V = m d^2 / (e t^2) = (9.1 \times 10^{-31} \times (0.02)^2) / (1.6 \times 10^{-19} \times (10^{-9})^2)$ **$V \approx 2.28 \times 10^3$ V $\approx 2275$ V**

Narrow electron beam deflected between two parallel horizontal plates

Electric Charges and Fields

Deflection of an electron beam in a uniform electric field

Official Answer

Data & setup

Entry speed (horizontal) $v = 3 \times 10^7$ m/s; plate length $l = 3\text{ cm} = 0.03\text{ m}$ ; separation $d = 2\text{ cm} = 0.02\text{ m}$ .
Entering midway, the beam must deflect $y = d/2 = 0.01$ m to just hit end $P_2'$ .
$m = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C.

Motion (like a projectile)

Time between plates: $t = l/v = 0.03 / (3 \times 10^7) = 1 \times 10^{-9}$ s.
Vertical acceleration: $a = eE/m = eV/(m d)$ .
Deflection: $y = \frac{1}{2} a t^2 = \frac{1}{2} \cdot eV/(md) \cdot t^2 = d/2$ .

Solve for V

$V = m d^2 / (e t^2) = (9.1 \times 10^{-31} \times (0.02)^2) / (1.6 \times 10^{-19} \times (10^{-9})^2)$

$V \approx 2.28 \times 10^3$ V $\approx 2275$ V

electron deflectionparallel platesuniform electric fieldprojectile motiontransit time l/vy = d/2V = m d^2/(e t^2)2275 V

Marking Scheme

11 mark: recognising projectile motion — $t = l/v$ and required deflection $y = d/2$ .
21 mark: correct equation $\frac{1}{2}\cdot(eV/md)\cdot t^2 = d/2$ leading to $V = m d^2/(e t^2)$ .
31 mark: correct substitution and final value $V \approx 2275$ V ( $\approx 2.28 \times 10^3$ V).

Hint

It is projectile motion: $t = l/v$ horizontally, and vertical drop $y = \frac{1}{2}(eV/md)t^2$ must equal $d/2$ (beam enters at the centre).

Quick Oral Answer

The electron crosses the 3 cm plates in a nanosecond while the field pulls it sideways like a projectile; setting the $\frac{1}{2}at^2$ drop equal to half the 2 cm gap gives $V = m d^2/e t^2 \approx 2275$ volts.

Analysis & Explanation

Concept

Inside parallel plates the field is uniform, so a charged particle moves exactly like a horizontal projectile: constant horizontal velocity and constant vertical acceleration $a = eE/m$ . This is the working principle of a cathode-ray oscilloscope deflection system.

Key modelling choices

The beam enters midway, so 'just striking the far end $P_2'$ ' means the vertical drop equals half the gap, $y = d/2 = 1$ cm.
The horizontal length l fixes the transit time $t = l/v$ ; the field acts only during this time.

Exam trap

Using $y = d$ (full gap) instead of $y = d/2$ doubles the answer — remember the beam starts at the centre.
Mixing cm and m; keep everything in SI. Watch that $t^2 = 10^{-18}\ \text{s}^2$ , a very small number.

Real-world link

Exactly this deflection physics steered the electron beam in old CRT televisions and oscilloscopes; the same E-field-vs-transit-time balance governs ink-jet printer droplet steering and mass-spectrometer beam control.

Common Mistakes

1Taking the required deflection as the full gap $d$ instead of $d/2$ , because the beam enters symmetrically (midway) — this doubles the voltage.
2Forgetting to convert cm to metres, giving answers off by powers of ten.
3Using the plate separation instead of the plate length to find the transit time $t = l/v$ .

Interesting Facts

This is precisely how a cathode-ray oscilloscope works: a voltage on the deflecting plates steers the electron beam, and because electrons are so light they respond within nanoseconds.

In J. J. Thomson's 1897 experiment the very same plate-deflection geometry, combined with a magnetic field, first yielded the electron's charge-to-mass ratio e/m.

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Frequently Asked Questions

Why is the required vertical deflection $d/2$ and not $d$ ?

The beam enters exactly midway between the plates (symmetrically). To 'just strike the far end $P_2'$ ' — the edge of the lower plate — it needs to drop only half the plate separation, i.e. $y = d/2 = 1$ cm, not the full 2 cm gap.

Does the electron's entry speed appear in the final formula?

Yes, through the transit time. A faster electron spends less time ( $t = l/v$ ) in the field, so it deflects less; to still hit the end you would need a larger voltage. Here $v = 3\times10^7$ m/s gives $t = 1$ ns and $V \approx 2275$ V.