Q1 · Electrostatics

Problem

Two large vertical parallel plates, separation 1 cm, potential difference X. A proton released at rest midway moves at $45°$ to the vertical just after release. Find X.

Options

Option	Value
(A)	$1\times 10^{-5}\,\text{V}$
(B)	$1\times 10^{-7}\,\text{V}$
(C)	$1\times 10^{-9}\,\text{V}$
(D)	$1\times 10^{-10}\,\text{V}$

Key insight: When a charged particle moves at 45° to the vertical, the electric and gravitational forces are equal in magnitude — this is a force diagram problem, not kinematics.

Setup

flowchart TB
  subgraph setup["Parallel plates, d = 1 cm"]
    P["Proton released at midpoint"]
  end
  subgraph forces["Forces at t = 0"]
    E["Electric: F_e = qE →"]
    G["Weight: F_g = mg ↓"]
  end
  P --> E
  P --> G
  E --> C["|F_e| = |F_g| ⇒ launch at 45°"]
  G --> C

1. Draw forces at release

Electric force F_e = qE (horizontal toward negative plate). Weight F_g = mg (downward). Particle starts from rest so acceleration direction = resultant of these two.

$$\vec{F} = qE\,\hat{x} - mg\,\hat{y}$$

2. 45° condition

For 45° below horizontal (or to vertical per problem statement), |qE| = mg.

$$qE = mg$$

3. Field from voltage

Parallel plates: E = X/d with d = 0.01 m.

$$E = X / 0.01$$

4. Numerical answer

X = mgd/q = (1.67×10⁻²⁷ × 10 × 0.01)/(1.6×10⁻¹⁹) ≈ 10⁻⁹ V.

$$X \approx 10^{-9}\,\text{V \;(C)}$$

Deep dive

A proton in a uniform vertical E-field between parallel plates experiences qE horizontally and mg downward. The instant it is released, its acceleration direction is tan⁻¹(qE/mg). Setting this to 45° gives qE = mg. The potential difference X = E·d where d = 1 cm.

Common pitfalls

• Using kinematics equations instead of force balance at t=0
• Forgetting d = 1 cm when converting E to X
• Sign errors on proton charge

Hints

• At 45°, electric force equals weight: qE = mg.

Answer

(C)