A particle of mass 10^-3 kg and charge 1.0 C is initially at rest. At time t = 0, the particle comes

Q: Why do we use $F = qE$ instead of Coulomb's law?

Coulomb's law is used to find the force between two point charges. When the electric field $\mathbf{E}$ is already given, the force on a charge $q$ is simply $F = qE$, which is a direct consequence of the definition of the electric field.

Q: Why did we integrate from $0$ to $v$ and $0$ to $t$?

The problem states that the particle is "initially at rest", which means at time $t = 0$, its velocity $v = 0$. These serve as our lower limits for the definite integral, allowing us to find the exact velocity $v$ at any arbitrary time $t$.

Q: What happened to the $10^3$ outside the integral?

When we integrated $\sin(10^3 t)$, we had to use substitution (or the reverse chain rule), which brought a factor of $\frac{1}{10^3}$ out of the integral. This perfectly canceled the $10^3$ that was already outside, leaving us with just the cosine term.

Q: Why is the maximum speed when $\cos(10^3 t) = -1$?

Our velocity function is $v(t) = 1 - \cos(10^3 t)$. To maximize $v(t)$, we need to subtract the smallest possible number from $1$. Since the range of the cosine function is $[-1, 1]$, its minimum value is $-1$. Subtracting $-1$ gives $1 - (-1) = 2$.

Q: Does the particle ever move backwards?

No! Because the velocity function is $v(t) = 1 - \cos(10^3 t)$, and the cosine function never exceeds $1$, the velocity is always greater than or equal to zero. The particle stops momentarily but never reverses its direction.

Question

A particle of mass $10^{-3}	ext{ kg}$ and charge $1.0	ext{ C}$ is initially at rest. At time $t = 0$, the particle comes under the influence of an electric field $\mathbf{E}(t) = E_0 \sin \omega t \hat{\mathbf{i}}$, where $E_0 = 1.0	ext{ N C}^{-1}$ and $\omega = 10^3	ext{ rad s}^{-1}$. Consider the effect of only the electrical force on the particle. Then, the maximum speed in $	ext{m s}^{-1}$, attained by the particle at subsequent times is ......

Sigma Percentile · Accepted Answer

### The Oscillating Electric Field

Imagine a charged particle sitting perfectly still on the x-axis. Suddenly, an electric field switches on, but it's not constant—it's oscillating like a sine wave!

This means the force on the particle is constantly changing in magnitude and direction. To understand how this affects the particle's motion, we need to start with the fundamental laws of physics.

### Newton Meets Lorentz

The electric field exerts a force on the particle, and according to Newton's second law, this force causes acceleration. We can write this relationship as:

$$F = ma = qE$$

We can express the acceleration as the rate of change of velocity, $\frac{dv}{dt}$. Substituting the given expression for the electric field, we get:

$$m \frac{dv}{dt} = q E_0 \sin(\omega t)$$

Let's isolate the acceleration by dividing both sides by the mass. This gives us the raw equation for how the particle's velocity changes over time:

$$a = \frac{dv}{dt} = \frac{q E_0}{m} \sin(\omega t)$$

### Crunching the Numbers

Now, let's plug in the given numerical values. The charge $q$ is $1.0 	ext{ C}$, the field amplitude $E_0$ is $1.0 	ext{ N/C}$, and the mass $m$ is $10^{-3} 	ext{ kg}$.

$$\frac{q E_0}{m} = \frac{1.0 	imes 1.0}{10^{-3}} = 10^3$$

Substituting this back, our differential equation becomes beautifully simple:

$$\frac{dv}{dt} = 10^3 \sin(10^3 t)$$

### Integrating for Velocity

To find the actual velocity at any time $t$, we need to work backwards from acceleration. We do this by integrating the acceleration with respect to time. We set up our definite integrals, knowing the particle starts from rest:

$$\int_0^v dv = \int_0^t 10^3 \sin(10^3 t) dt$$

The integral of sine is negative cosine. Applying the reverse chain rule, we divide by the coefficient of $t$, which is $10^3$. Notice how the thousands cancel out perfectly!

$$v(t) = 10^3 \left[ \frac{-\cos(10^3 t)}{10^3} ight]_0^t$$

Applying the upper and lower limits, we get:

$$v(t) = - (\cos(10^3 t) - \cos(0))$$

Since $\cos(0) = 1$, the velocity function simplifies to:

$$v(t) = 1 - \cos(10^3 t)$$

### Finding the Maximum Speed

We are looking for the maximum speed. In our velocity equation, we are subtracting a cosine term. To make the velocity as large as possible, we need to subtract the smallest possible value.

The minimum value of a cosine function is $-1$. This happens when the electric field has completely reversed its direction and pushed the particle to its peak speed.

Substituting $-1$ into our equation, we get:

$$v_{\max} = 1 - (-1) = 2 	ext{ m/s}$$

**The maximum speed attained by the particle is 2 m/s.**

### The Way Forward

Think about this: what if the electric field was a cosine wave instead of a sine wave? The velocity would then be a sine wave, meaning the particle would oscillate back and forth, crossing the origin repeatedly! The initial phase of the field completely changes the macroscopic trajectory of the particle.

Enter Numerical Value:

Visualized Solution

$Visual Anchor$

$Newton \overset{s}{ˊ} Second Law$

$Equation of Motion$

$Isolating Acceleration$

$Substituting Values$

$Differential Equation$

$Integration Setup$

$Definite Integrals$

$Performing Integration$

$Velocity Function$

$Maximizing Velocity$

$Final Answer$

$The Way Forward$

The Sigma Insight: Electric Field

The Oscillating Electric Field

Newton Meets Lorentz

Crunching the Numbers

Integrating for Velocity

Finding the Maximum Speed

The Way Forward

Similar Questions

A particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

A charged particle (mass $m$ and charge $q$ ) moves along X-axis with velocity $v_{0}$ . When it passes through the origin it enters a region having uniform electric field $E = - E \hat{j}$ which extends upto $x = d$ . Equation of path of electron in the region ( $x > d$ ) is

A positive point charge is released from rest at a distance $r_{0}$ from a positive line charge with uniform density. The speed ( $v$ ) of the point charge, as a function of instantaneous distance $r$ from line charge, is proportional to

Two point charges $q_{1} (10 μ C)$ and $q_{2} (- 25 μ C)$ are placed on the x-axis at $x = 1 m$ and $x = 4 m$ , respectively. The electric field (in V/m) at a point $y = 3 m$ on Y-axis is $(Take, \frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} N-m^{2} C^{- 2})$

Three charged particles $A, B$ and $C$ with charges $- 4 q, 2 q$ and $- 2 q$ are present on the circumference of a circle of radius $d$ . The charged particles $A, C$ and centre $O$ of the circle formed an equilateral triangle as shown in figure. Electric field at $O$ along x-direction is

The bob of a simple pendulum has mass $2 g$ and a charge of $5.0 μ C$ . It is at rest in a uniform horizontal electric field of intensity $2000 V/m$ . At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10 m/s^{2}$ )

Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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

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

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

.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