Q.An electron falls through a distance of $1.5\,\text{cm}$ in a uniform electric field of magnitude $2.0 \times 10^{4}\,\text{N C}^{-1}$ [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

In a uniform field the electric force $F=qE$ is constant, so each particle has constant acceleration $a=qE/m$ and, starting from rest, falls through $s$ in $t=\sqrt{2s/a}$.

Data: $s=1.5\,\text{cm}=0.015\,\text{m}$, $E=2.0\times10^{4}\,\text{N/C}$, $e=1.6\times10^{-19}\,\text{C}$.

Electron ($m_e=9.11\times10^{-31}\,\text{kg}$):

Proton ($m_p=1.67\times10^{-27}\,\text{kg}$):

Contrast: in gravity $a=g\approx9.8\,\text{m/s}^2$ is the same for every body, whereas here $a=qE/m$ depends on the charge-to-mass ratio, so the light electron falls much faster than the proton, and both times are far shorter than a gravitational fall ($\approx0.055\,\text{s}$).

A uniform field exerts a constant force $F=qE$, hence a constant acceleration by Newton's second law. Starting from rest, kinematics gives $s=\tfrac12 a t^2$, so $t=\sqrt{2s/a}$.

Given: $s=1.5\,\text{cm}=0.015\,\text{m}$, $E=2.0\times10^{4}\,\text{N/C}$, $e=1.6\times10^{-19}\,\text{C}$, $m_e=9.11\times10^{-31}\,\text{kg}$, $m_p=1.67\times10^{-27}\,\text{kg}$. In (a) the field points up and the electron (negative) is pushed down; in (b) the field is reversed and the proton (positive) is also pushed down — so each falls through the same $s$.

Electron.

Proton. The force magnitude $eE=3.2\times10^{-15}\,\text{N}$ is the same, but the proton is about $1836$ times heavier:

Contrast with free fall under gravity. In free fall the acceleration is $g\approx9.8\,\text{m/s}^2$, identical for every body regardless of mass. Here the acceleration $a=qE/m$ is enormous ($\sim10^{12}$–$10^{15}\,\text{m/s}^2$) and depends on the charge‑to‑mass ratio, so electron and proton fall in very different times. A purely gravitational fall through the same $0.015\,\text{m}$ would take $t_g=\sqrt{2s/g}\approx0.055\,\text{s}$ — millions of times longer, which is why gravity is negligible for charged particles in such fields.

Method: Newton’s Second Law + Kinematics (Uniform Acceleration)

This method uses the electric force to find acceleration, then applies equations of motion for constant acceleration.

Step 1 — Identify the force and acceleration

For a charge $q$ in a uniform electric field $\mathbf{E}$:

By Newton’s second law:

• For the electron: $q = -e$, so $a_e = \frac{-eE}{m_e}$ (magnitude $a_e = \frac{eE}{m_e}$)
• For the proton: $q = +e$, so $a_p = \frac{eE}{m_p}$

Step 2 — Apply kinematics for constant acceleration

The particle falls from rest ($u = 0$) through distance $s = 1.5\ \text{cm} = 0.015\ \text{m}$.

Using $s = ut + \frac{1}{2} a t^2$:

Step 3 — Compute times

Given:

$E = 2.0 \times 10^4\ \text{N C}^{-1}$

$e = 1.6 \times 10^{-19}\ \text{C}$

$m_e = 9.1 \times 10^{-31}\ \text{kg}$

$m_p = 1.67 \times 10^{-27}\ \text{kg}$

For electron:

For proton:

Step 4 — Contrast with free fall under gravity

• Free fall acceleration: $g \approx 9.8\ \text{m s}^{-2}$ Time to fall $1.5\ \text{cm}$:

• Key differences:
  • Electric acceleration is huge compared to $g$ (by factors of $10^{11}$ to $10^{14}$)
  • Electron falls ~43 times faster than proton (because $m_e \ll m_p$)
  • In free fall, all objects fall with same acceleration (independent of mass) — here, acceleration depends on charge-to-mass ratio $q/m$

Final results:

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the Mass Difference Between Electron and Proton

The error: Students often assume both particles have the same mass, or they plug in the electron mass for the proton (or vice versa).

Why it happens: The problem asks for "time of fall" for both, and it's easy to rush and use the same mass value.

How to avoid:

• Always write the mass explicitly before substituting:
  • Electron: $m_e = 9.1 \times 10^{-31}\,\text{kg}$
  • Proton: $m_p = 1.67 \times 10^{-27}\,\text{kg}$
• Note that the proton is ~1836 times heavier — this alone tells you the proton will take much longer to fall the same distance.

Mistake 2: Ignoring the Sign of Charge When Field Reverses

The error: Students treat the reversed field as if it still pushes the proton in the same direction as the electron.

Why it happens: The problem says "direction of the field is reversed" — but the proton is positively charged, so the force direction flips relative to the electron.

How to avoid:

• Force on a charge: $\vec{F} = q\vec{E}$
  • Electron ($q = -e$): force is opposite to $\vec{E}$
  • Proton ($q = +e$): force is along $\vec{E}$
• When the field reverses, the electron would reverse direction too — but here the electron falls before reversal, and the proton falls after reversal.
• Key insight: In both cases, the particle is accelerated downward (toward the lower plate). The reversal ensures the proton also falls, not rises.

Mistake 3: Using $g$ Instead of Electric Acceleration

The error: Students treat this as free fall under gravity and use $t = \sqrt{2h/g}$.

Why it happens: The problem explicitly asks to "contrast with free fall under gravity," but students sometimes mix the two.

How to avoid:

• The acceleration here is NOT $g$. It comes from the electric force:

• For the electron: $a_e = \frac{eE}{m_e}$
• For the proton: $a_p = \frac{eE}{m_p}$
• Then use $s = \frac{1}{2} a t^2$ → $t = \sqrt{\frac{2s}{a}}$

Mistake 4: Forgetting That Both Charges Have Magnitude $e$

The error: Students think the proton has charge $+e$ and the electron $-e$, but then use different magnitudes of charge in the force equation.

Why it happens: The sign matters for direction, but the magnitude of charge is the same: $e = 1.6 \times 10^{-19}\,\text{C}$.

How to avoid:

• Write: $|q_e| = |q_p| = e$
• The force magnitude is $eE$ for both — only the mass differs.

Mistake 5: Not Contrasting with Free Fall Properly

The error: Students compute the times but don't explain why the comparison matters.

Why it happens: The question says "Contrast the situation," but students treat it as an afterthought.

How to avoid:

• Free fall under gravity:

• Electric case:

• Key contrast:
  • In free fall, all objects fall with same acceleration $g$ (ignoring air resistance).
  • In an electric field, acceleration depends on mass — the electron falls ~1836× faster than the proton.
  • Also, gravity always pulls downward; electric force direction depends on charge sign.

Quick Checklist to Avoid These Mistakes

Final Answer (for reference)

Electron:

Proton:

Contrast: Under gravity, both would take $\sqrt{2h/g} \approx 0.055\,\text{s}$ — much slower than the electron, but faster than the proton. The electric field discriminates by mass, while gravity does not.