A particle of charge q and mass m is subjected to an electric field E = E_0(1 - ax^2) in the x-direc

Q: Why can't we use the equation $v^2 = u^2 + 2as$?

That equation is strictly for constant acceleration. Here, the electric field (and thus the force and acceleration) depends on $x$, meaning it is variable. You must use integration or energy methods.

Q: Why is the change in kinetic energy zero?

The particle starts from rest ($v=0$), so its initial kinetic energy is zero. The question asks for the point where it comes to rest again, so its final kinetic energy is also zero. Thus, $0 - 0 = 0$.

Q: What happens to the particle after it reaches $x_0$?

At $x_0$, the particle stops momentarily. Because the electric field at $x_0$ is negative, it will experience a force pushing it back towards the origin. It will actually oscillate back and forth!

Q: Why did we ignore the $x_0 = 0$ solution?

The math correctly identifies that the kinetic energy is zero at $x=0$ (the starting point). However, the question specifically asks for the position *other than the initial position* where the kinetic energy becomes zero.

Q: Can we solve this by finding the potential energy function?

Absolutely! The work done by the electric field is the negative change in potential energy. Setting $\Delta U = 0$ will give you the exact same integral and the exact same result.

Question

A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_0(1 - ax^2)$ in the x-direction, where $a$ and $E_0$ are constants. Initially, the particle was at rest at $x = 0$. Other than the initial position, the kinetic energy of the particle becomes zero when the distance of the particle from the origin is

Options:

(A) $\sqrt{\frac{1}{a}}$

(B) $a$

(C) $\sqrt{\frac{3}{a}}$

(D) $\sqrt{\frac{2}{a}}$

Sigma Percentile · Accepted Answer

The journey of a charged particle through a varying electric field is a classic exploration of energy and motion. In this problem, we are tasked with finding the exact point where a particle, initially at rest, comes to a halt once again.

### Analyzing the Setup

Imagine you are observing a particle of mass $m$ and charge $q$ placed at the origin. The electric field it experiences is not constant; it is given by the equation:

$$E = E_0(1 - ax^2)$$

Initially, at $x = 0$, the field is positive ($E_0$), which means the particle experiences a force pushing it in the positive x-direction. As it moves, the field strength decreases, eventually becoming zero at $x = \frac{1}{\sqrt{a}}$. Beyond this point, the field becomes negative, acting as a restoring force that slows the particle down.

### The Master Equation

To find where the particle stops, we could try using Newton's laws to find acceleration and then integrate to find velocity. However, there is a much more elegant approach: the **Work-Energy Theorem**. This theorem states that the net work done on an object is equal to its change in kinetic energy:

$$W_{net} = \Delta K$$

Since the particle starts from rest, its initial kinetic energy $K_i$ is zero. We are looking for the position $x_0$ where it comes to rest again, meaning its final kinetic energy $K_f$ is also zero. Therefore, the total change in kinetic energy is zero:

$$\Delta K = 0 \implies W_{net} = 0$$

### The Integration Phase

The work done by a variable force is the integral of the force over the displacement. The electric force is $F = qE$. Setting the total work to zero gives us:

$$\int_{0}^{x_0} qE \, dx = 0$$

Substituting our specific electric field into the integral:

$$\int_{0}^{x_0} q \left[ E_0(1 - ax^2) \right] dx = 0$$

Since $q$ and $E_0$ are non-zero constants, we can divide them out. We are left with a straightforward polynomial integration:

$$\left[ x - \frac{ax^3}{3} \right]_{0}^{x_0} = 0$$

### Final Calculation

Applying the limits from $0$ to $x_0$, we get:

$$x_0 - \frac{ax_0^3}{3} = 0$$

We can factor out $x_0$:

$$x_0 \left( 1 - \frac{ax_0^2}{3} \right) = 0$$

This equation has two solutions. The first is $x_0 = 0$, which is our starting point. The second solution, which is the one we are looking for, occurs when the term in the parentheses is zero:

$$1 = \frac{ax_0^2}{3}$$

Solving for $x_0$, we find:

$$x_0^2 = \frac{3}{a} \implies x_0 = \sqrt{\frac{3}{a}}$$

This is the exact distance from the origin where the positive work done by the initial forward push is perfectly canceled out by the negative work done by the opposing field, bringing the particle to a momentary halt.

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Visualized Solution

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$A ppl y in g t h e T h eor e m$

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The Sigma Insight: Work Done by Forces

Analyzing the Setup

The Master Equation

The Integration Phase

Final Calculation

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Visualized Solution

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