The specific charge of electron is (A) 1.8 × 10^-19 C/kg (B) 1.67 × 10^-19 C/kg (C) 1.8 × 10¹1 C/kg (D) 6.67 ×

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Q. The specific charge of electron is (A) 1.8 \times 10^-19 C/kg (B) 1.67 \times 10^-19 C/kg (C) 1.8 \times 10^11 C/kg (D) 6.67 \times 10^11 C/kg

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Concept understanding — Charge To Mass Ratio

Charge to Mass Ratio: The First Meeting

Imagine you have two identical-looking balls. One is made of cork, the other of lead. If you blow on them with the same fan, the cork ball flies away easily, while the lead ball barely moves. The difference isn't in the force you applied — it's in how much mass each ball has. Now imagine that instead of blowing air, you're using an electric field to push a charged particle. The same idea applies: the particle's response depends on both its charge (how strongly the field pushes it) and its mass (how much inertia it has to overcome).

The charge to mass ratio ( $e/m$ for an electron, often written as $q/m$ in general) is simply the charge of a particle divided by its mass. It tells you how much "electric responsiveness" a particle has per unit of its inertia.

Why This Ratio Matters

Two particles can have the same charge but very different masses. A proton and a positron both have charge $+e$ , but the proton is about 1836 times heavier. If you put them in the same electric field, the positron accelerates 1836 times more. The charge-to-mass ratio captures this difference in a single number.

For an electron:

\frac{e}{m_e} \approx 1.76 \times 10^{11} \, \text{C/kg}

For a proton:

\frac{e}{m_p} \approx 9.58 \times 10^{7} \, \text{C/kg}

The electron's ratio is nearly 2000 times larger. That's why electrons are so much more mobile in circuits and beams — they respond far more dramatically to electric and magnetic fields.

The Precise Definition

\frac{q}{m} = \frac{\text{charge of the particle}}{\text{mass of the particle}}

Its SI unit is coulombs per kilogram (C/kg). For any charged particle, this ratio determines:

How much it accelerates in an electric field: $a = \frac{qE}{m}$
How tightly it curves in a magnetic field: radius $r = \frac{mv}{qB}$
The frequency of its circular motion: $\omega = \frac{qB}{m}$

How It Was Discovered

J.J. Thomson measured this ratio for cathode rays in 1897. He didn't know what the particles were — he just knew they were charged and had mass. By balancing electric and magnetic forces, he found that the ratio was over a thousand times larger than for any known ion. This told him the particles (which we now call electrons) were either extraordinarily charged or extraordinarily light. We now know it's the latter: the electron is the lightest charged particle with a non-zero rest mass.

Note

The charge-to-mass ratio is a fundamental property of each type of particle. It cannot be changed — it's as intrinsic as the particle's spin or its rest mass. …

Why this formula?

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

The charge-to-mass ratio ( $e/m$ ) is a fundamental property of charged particles. For the electron, its measurement was a landmark experiment by J.J. Thomson (1897). Let's understand why the key formula emerges from the physics.

1. The Core Idea: Balancing Forces

The experiment uses a velocity selector — a region with perpendicular electric ( $\vec{E}$ ) and magnetic ( $\vec{B}$ ) fields.

What happens to a charged particle?

A particle with charge $q$ and mass $m$ moving with velocity $\vec{v}$ experiences:

Electric force: $\vec{F}_E = q\vec{E}$ (direction: along $\vec{E}$ )
Magnetic force: $\vec{F}_B = q(\vec{v} \times \vec{B})$ (direction: perpendicular to both $\vec{v}$ and $\vec{B}$ )

The key insight:

If we arrange $\vec{E}$ and $\vec{B}$ perpendicular to each other and to $\vec{v}$ , the two forces act in opposite directions.

2. Deriving the Velocity Condition

For the particle to pass undeflected through the crossed fields:

\vec{F}_E + \vec{F}_B = 0

Since forces are opposite:

qE = qvB

Cancel $q$ (non-zero for a charged particle):

\boxed{v = \frac{E}{B}}

Why this matters: This gives us the particle's speed without knowing its mass or charge. The velocity selector picks out only particles with this specific speed.

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

After the velocity selector, the particle enters a region with only magnetic field ( $B$ ). Here:

Magnetic force provides centripetal force
The particle moves in a circular path of radius $r$

Force balance:

qvB = \frac{mv^2}{r}

Rearranging:

\frac{q}{m} = \frac{v}{Br}

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

\boxed{\frac{q}{m} = \frac{E}{B^2 r}}

4. Why This Formula Holds — The Physical Logic

Step	Physics Principle	What it gives us
1	Force balance in crossed fields	Speed $v = E/B$

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Specific charge is the charge-to-mass ratio e/m ≈ 1.76×10¹¹ C/kg, i.e. about 1.8×10¹¹ C/kg.

The specific charge of the electron is its charge divided by its mass:

$\frac{e}{m} = \frac{1.6\times10^{-19}\,C}{9.1\times10^{-31}\,kg} \approx 1.76\times10^{11}\,C/kg$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set D1 markMCQ
$Q. The specific charge of electron is (A) 1.8 \times 10^-19 C/kg (B) 1.67 \times 10^-19 C/kg (C) 1.8 \times 10^11 C/kg (D) 6.67 \times 10^11 C/kg$
›Reveal solutionSolution

Specific charge is the charge-to-mass ratio e/m ≈ 1.76×10¹¹ C/kg, i.e. about 1.8×10¹¹ C/kg.

The specific charge of the electron is its charge divided by its mass:

$\frac{e}{m} = \frac{1.6\times10^{-19}\,C}{9.1\times10^{-31}\,kg} \approx 1.76\times10^{11}\,C/kg$

…

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CBSE 2024Set 55/2/11 markMCQ
$m_P$
›Reveal solutionSolution
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

Write the force on each particle. Both have charge $q$ (same magnitude), so:

$F_P = qE, \quad F_Q = qE$

Find the acceleration of each particle. Using $F = ma$ :

$a_P = \frac{qE}{m_P}, \quad a_Q = \frac{qE}{m_Q}$

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:
$s = \frac{1}{2}a_Q t_Q^2 = \frac{1}{2} \cdot \frac{qE}{m_Q} \cdot t_Q^2$

Equate the two expressions for $s$ . Since both travel the same distance: …
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CBSE 2020Set HE8211 mark
$Q. Fill in the blank: The value of specific charge of electron in S.I. unit is ______ C/kg.$
›Reveal solutionSolution

The specific charge (charge-to-mass ratio) of the electron is 1.76 × 10¹¹ C/kg.

…

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Q.The specific charge of electron is (A) 1.8 × 10^-19 C/kg (B) 1.67 × 10^-19 C/kg (C) 1.8 × 10^11 C/kg (D) 6.67 × 10^11 C/kg

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Q.The specific charge of electron is (A) 1.8 × 10^-19 C/kg
(B) 1.67 × 10^-19 C/kg
(C) 1.8 × 10^11 C/kg
(D) 6.67 × 10^11 C/kg

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