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An electron fired between two charged plates bends the way a ball thrown sideways falls: steady speed along the plates, steady acceleration across them. Once you see that, every deflection problem is a projectile problem with a different acceleration. This post derives the deflection, works one full example with a screen, and lists the traps.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "force on a moving charge in uniform magnetic and electric fields" in Unit 13, and the CBSE Physics curriculum for 2026-27 has the same line in Class XII Chapter 4. The NCERT chapter Electric Charges and Fields works the simplest case in Example 1.7: an electron falling 1.5 cm from rest in a field of 2.0 × 10⁴ N/C takes 2.9 × 10⁻⁹ s, and NCERT shows that gravity can be ignored because the electric acceleration is enormous by comparison. Check your own exam year's syllabus.
Why is the path a parabola?
Take plates of length L with a uniform field E between them, and an electron entering midway with speed v₀ parallel to the plates. Ignore edge effects.
Along the plates there is no force, so x = v₀t.
Across the plates the force is eE, so the acceleration is a = eE/m, constant, towards the positive plate.
So y = ½at² = (eE / 2mv₀²) x². That is the same shape as a ball thrown horizontally, y = (g / 2u²) x², with eE/m in place of g.
At the exit, t = L/v₀, which gives:
deflection y₁ = eEL² / (2mv₀²)
sideways speed v_y = eEL / (mv₀)
exit angle tan θ = eEL / (mv₀²)
Beyond the plates there is no field, so the electron moves in a straight line. Extend that line backwards and it meets the axis at the centre of the plates, because tan θ × L/2 = y₁. On a screen a distance D beyond the plates, the total deflection is Y = tan θ × (L/2 + D).
Worked example: an electron reaching a screen
E = 5.0 × 10³ N/C (100 V across a 2.0 cm gap), L = 4.0 cm, v₀ = 2.0 × 10⁷ m/s, and a screen 20 cm beyond the plates. Use e = 1.602 × 10⁻¹⁹ C and m = 9.109 × 10⁻³¹ kg (NIST values for e and the electron mass).
Acceleration: a = eE/m = (1.602 × 10⁻¹⁹ × 5.0 × 10³) / 9.109 × 10⁻³¹ ≈ 8.79 × 10¹⁴ m/s².
Time between the plates: t = 0.040 / 2.0 × 10⁷ = 2.0 × 10⁻⁹ s.
Deflection at exit: y₁ = ½ × 8.79 × 10¹⁴ × (2.0 × 10⁻⁹)² ≈ 1.76 × 10⁻³ m, or 1.76 mm. The electron entered 1.0 cm from each plate, so it clears them.
Sideways speed: v_y = at ≈ 1.76 × 10⁶ m/s, so tan θ ≈ 1.76 × 10⁶ / 2.0 × 10⁷ ≈ 0.088 and θ ≈ 5.0°.
On the screen: Y = 0.088 × (0.020 + 0.20) ≈ 0.0193 m, about 19 mm.
The acceleration is roughly 9 × 10¹³ times g, so gravity really is negligible. The speed is about 7% of the speed of light, so the non-relativistic formulas are fine at this level.
What changes for other particles and speeds?
The deflection is y₁ = qEL² / (2mv₀²) = qEL² / (4K), where K is the kinetic energy. Two consequences:
Faster or heavier means less deflection. Doubling v₀ cuts the deflection to a quarter. A proton at the same speed, using NCERT's masses (1.67 × 10⁻²⁷ kg against 9.11 × 10⁻³¹ kg), is about 1,800 times heavier and deflects by roughly a micrometre.
Same accelerating voltage, same path. If the particle was accelerated from rest through a voltage V_a, then K = qV_a and y₁ = EL² / (4V_a). The charge and mass cancel. In our example the electron's energy corresponds to V_a ≈ 1.14 kV, and EL²/(4V_a) gives the same 1.76 mm.
How did this experiment start?
In his 1897 paper Cathode Rays, J. J. Thomson sent cathode rays between two parallel aluminium plates about 5 cm long and 1.5 cm apart, and reported that the deflection was proportional to the potential difference between the plates. Heinrich Hertz had earlier seen no deflection, and Thomson traced that to the gas in the tube conducting. Our profile of J. J. Thomson and the electron tells the rest of the story.
Which mistakes cost marks?
Using the total time to the screen in y = ½at². The acceleration acts only between the plates.
Keeping the path curved after the plates. It is straight there.
Getting the direction wrong. An electron accelerates towards the positive plate, opposite to E.
Using V/L for the field. The field is V/d, where d is the gap between the plates.
Expecting faster electrons to deflect more. Deflection falls as 1/v₀².
A practice routine
Day 1: Derive y₁, tan θ and Y from scratch, side by side with the projectile formulas.
Day 2: Redo the worked example with v₀ = 4.0 × 10⁷ m/s and check that both deflections fall to a quarter.
Day 3: Solve two problems that give the accelerating voltage instead of the speed, using y₁ = EL²/(4V_a).
Day 4: Mix in two charged-particle problems in magnetic fields and compare the paths.
Day 5: Do a mixed test without notes. Dunlosky and colleagues (2013) rated practice testing as one of the two most useful study techniques.
If you find the path hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, sketch the electron's path in the worked example to scale, with the straight section traced back to the middle of the plates.
