Two charged particles P and Q, having the same charge but different masses $m_P$ and $m_Q$, start from rest and travel equal distances in a uniform electric field $\vec{E}$ in time $t_P$ and $t_Q$ respectively. Neglecting the effect of gravity, the ratio $\dfrac{t_P}{t_Q}$ is : (

Question

Two charged particles P and Q, having the same charge but different masses $m_P$ and $m_Q$, start from rest and travel equal distances in a uniform electric field $\vec{E}$ in time $t_P$ and $t_Q$ respectively. Neglecting the effect of gravity, the ratio $\dfrac{t_P}{t_Q}$ is : (A) $\dfrac{m_P}{m_Q}$ (B) $\dfrac{m_Q}{m_P}$ (C) $\sqrt{\dfrac{m_P}{m_Q}}$ (D) $\sqrt{\dfrac{m_Q}{m_P}}$

Accepted Answer

> [!TLDR]
> Both particles experience the same force (same charge, same field) but different accelerations inversely proportional to their masses. Since distance $s = \frac{1}{2}at^2$ for motion from rest, time scales as $\sqrt{\frac{1}{a}} \propto \sqrt{m}$. The ratio is $\boxed{\sqrt{\dfrac{m_P}{m_Q}}}$.

## Why time depends on mass in a uniform field

When a charged particle moves in a uniform electric field, the force it experiences depends only on its charge: $F = qE$. But the resulting acceleration depends on mass through Newton's second law: $a = \frac{F}{m} = \frac{qE}{m}$.

Since both particles have the same charge $q$ and move in the same field $E$, they experience identical forces. The lighter particle accelerates more; the heavier one accelerates less. When they travel the same distance starting from rest, the one with smaller acceleration (larger mass) takes longer.

The key insight is that for motion under constant acceleration from rest, distance grows as the *square* of time. This means time grows as the *square root* of the inverse of acceleration—and hence as the square root of mass.

## Step-by-step solution

1. **Write the force on each particle.**  
   Both have charge $q$ (same magnitude), so:
   $$F_P = qE, \quad F_Q = qE$$

2. **Find the acceleration of each particle.**  
   Using $F = ma$:
   $$a_P = \frac{qE}{m_P}, \quad a_Q = \frac{qE}{m_Q}$$

3. **Apply the kinematic equation for distance.**  
   Both start from rest ($u = 0$) and travel the same distance $s$. The equation of motion is:
   $$s = ut + \frac{1}{2}at^2 = \frac{1}{2}at^2$$
   
   For particle P:
   $$s = \frac{1}{2}a_P t_P^2 = \frac{1}{2} \cdot \frac{qE}{m_P} \cdot t_P^2$$
   
   For particle Q:
   $$s = \frac{1}{2}a_Q t_Q^2 = \frac{1}{2} \cdot \frac{qE}{m_Q} \cdot t_Q^2$$

4. **Equate the two expressions for $s$.**  
   Since both travel the same distance:
   $$\frac{1}{2} \cdot \frac{qE}{m_P} \cdot t_P^2 = \frac{1}{2} \cdot \frac{qE}{m_Q} \cdot t_Q^2$$
   
   The factors $\

Step	Physics Principle	What it gives us
1	Force balance in crossed fields	Speed $v = E/B$
2	Centripetal force in magnetic field	Radius $r$ depends on $q/m$
3	Combine both	Direct measurement of $q/m$

Concept understanding — Charge To Mass Ratio

Charge to Mass Ratio: The First Meeting

Why This Ratio Matters

The Precise Definition

How It Was Discovered

A Common Confusion

The Takeaway

Why this formula?

Charge to Mass Ratio ( $e/m$ ) — Why the Formula Holds

1. The Core Idea: Balancing Forces

What happens to a charged particle?

The key insight:

2. Deriving the Velocity Condition

3. Measuring $e/m$ — The Circular Path

Force balance:

Rearranging:

Substituting $v = E/B$ from step 2:

4. Why This Formula Holds — The Physical Logic

Key assumptions (exam-relevant):

5. For the Electron: The Famous Result

Quick Exam Tip

Why time depends on mass in a uniform field

Step-by-step solution

Why time depends on mass in a uniform field

Step-by-step solution

Why time depends on mass in a uniform field

Step-by-step solution

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