An electron of mass m_e, initially at rest, moves through a certain distance in a uniform electric f

Q: Why do the electron and proton move in opposite directions?

The electric force on a charge is $\mathbf{F} = q\mathbf{E}$. Since the electron has a negative charge and the proton has a positive charge, the forces on them are in opposite directions.

Q: Why can we neglect gravity in this problem?

The electrostatic force between subatomic particles or on them in typical electric fields is many orders of magnitude ($>10^{30}$ times) stronger than the gravitational force.

Q: Does the initial position matter?

No, as long as the electric field is uniform, the acceleration is constant everywhere, so only the distance covered matters.

Q: What if they had an initial velocity?

If they had an initial velocity, the equation would be $S = ut + \frac{1}{2}at^2$, and the time ratio would depend on the initial velocity as well, making it a quadratic equation to solve.

Q: Is the time taken directly or inversely proportional to mass?

The time taken is directly proportional to the square root of the mass ($t \propto \sqrt{m}$). A heavier particle takes more time to cover the same distance because it accelerates slower.

Question

An electron of mass $m_e$, initially at rest, moves through a certain distance in a uniform electric field in time $t_1$. A proton of mass $m_p$, also, initially at rest, takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio $t_2/t_1$ is nearly equal to

Options:

(A) 1

(B) $(m_p/m_e)^{1/2}$

(C) $(m_e/m_p)^{1/2}$

(D) 1836

Sigma Percentile · Accepted Answer

### The Setup: A Subatomic Drag Race

Imagine a classic drag race, but on a subatomic scale. In one lane, we have an electron, incredibly light and nimble. In the other lane, a proton, much heavier but carrying an equal and opposite charge. The track is a uniform electric field $\mathbf{E}$, and both racers start from a complete standstill ($u = 0$).

Our goal is to find out how the time taken by the proton ($t_2$) compares to the time taken by the electron ($t_1$) to cover the exact same distance $S$.

### The Physics: Forces and Accelerations

When placed in an electric field, both particles experience an electrostatic force. The magnitude of this force is given by:

$$ F_e = eE $$

Even though the electron is negative and the proton is positive (meaning they will accelerate in opposite directions), the *magnitude* of the force pulling them is identical. However, Newton's Second Law ($F = ma$) tells us that acceleration depends on mass.

For the electron, the acceleration is:

$$ a_e = \frac{eE}{m_e} $$

For the proton, the acceleration is:

$$ a_p = \frac{eE}{m_p} $$

Because the proton is much heavier ($m_p \gg m_e$), its acceleration will be significantly smaller than that of the electron.

### The Math: Kinematics of Constant Acceleration

Since the electric field is uniform, the accelerations are constant. We can use the second equation of kinematics to relate distance, acceleration, and time:

$$ S = ut + \frac{1}{2}at^2 $$

Given that both particles start from rest ($u = 0$), the equation simplifies beautifully to:

$$ S = \frac{1}{2}at^2 $$

Because both particles cover the exact same distance $S$, we can equate their individual kinematic expressions:

$$ \frac{1}{2}a_e t_1^2 = \frac{1}{2}a_p t_2^2 $$

### The Finish Line: The Mass-Time Relationship

Now, it's just a matter of simple algebra. Let's rearrange the equation to find the ratio of their times squared:

$$ \frac{t_2^2}{t_1^2} = \frac{a_e}{a_p} $$

Substitute the expressions for accelerations we found earlier:

$$ \frac{t_2^2}{t_1^2} = \frac{\frac{eE}{m_e}}{\frac{eE}{m_p}} $$

The $eE$ terms cancel out perfectly, leaving us with the inverse ratio of their masses:

$$ \frac{t_2^2}{t_1^2} = \frac{m_p}{m_e} $$

Finally, taking the square root of both sides gives us the winning ratio:

$$ \frac{t_2}{t_1} = \sqrt{\frac{m_p}{m_e}} = \left(\frac{m_p}{m_e}\right)^{1/2} $$

**Conclusion:** The time taken to cover a specific distance from rest in a uniform electric field is directly proportional to the square root of the particle's mass. The heavier proton takes much longer to cross the finish line!

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Visualized Solution

$Initial Setup$

$Electrostatic Force & Acceleration$

$Equation of Kinematics$

$Equating Distances$

$Solving for Time Ratio$

$Final Answer$

$The Way Forward$

The Sigma Insight: Electric Field

The Setup

The Physics

The Math

The Finish Line

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electrostatic Force & Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Kinematics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Distances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Time Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward