Animated Solution for Physics - Electrostatics: A body having specific charge $8\mu \text{C/g}$ is resting on a frictionless plane at a distance $10\text{ cm}$ from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of $100\text{ V/m}$ is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ...... s.

Imagine a small charged body resting peacefully on a perfectly smooth, frictionless horizontal plane. Just $10\text{ cm}$ away stands a rigid wall. Suddenly, a uniform electric field of $100\text{ V/m}$ is switched on, pointing directly towards the wall. The body, possessing a specific charge of $8\mu \text{C/g}$, is immediately thrust into motion. Our goal is to determine the time period of this motion, given that the collision with the wall is perfectly elastic.

Before we dive into the kinematics, we must ensure all our units are speaking the same language. The specific charge is given in microcoulombs per gram. Let's convert this into standard SI units (Coulombs per kilogram):

Similarly, the distance to the wall is $s=10\text{ cm}=0.1\text{ m}$.

When a charge $q$ is placed in an electric field $E$, it experiences an electrostatic force given by $F=qE$. According to Newton's Second Law, this force will cause the body to accelerate ($F=ma$).

By equating the two expressions for force, we can find the acceleration:

Notice how beautifully the concept of specific charge ($q/m$) simplifies our work. We don't need to know the individual mass or charge of the body; their ratio is all that matters. Plugging in our values:

This acceleration is constant and directed horizontally towards the wall.

The Journey to the Wall

The body starts from rest ($u=0$) and accelerates uniformly at $0.8{\text{ m/s}}^2$ over a distance of $0.1\text{ m}$. We can use the second equation of kinematics to find the time $t$ it takes to reach the wall:

Since $u=0$, the equation simplifies to:

It takes exactly half a second for the body to strike the wall.

Here is where the physics gets truly interesting. The problem states that the collision with the wall is perfectly elastic. This means the coefficient of restitution $e=1$. The body rebounds with the exact same speed it had just before impact, but its velocity vector is now reversed, pointing away from the wall.

However, the electric field hasn't changed. It is still pushing the body towards the wall. Therefore, as the body moves away from the wall, the electric force acts as a restoring force, causing a constant deceleration of $-0.8{\text{ m/s}}^2$.

Kinematically, the process of decelerating from a velocity $v$ to $0$ under a constant acceleration $-a$ takes the exact same amount of time as accelerating from $0$ to $v$ under an acceleration $a$. Thus, the time taken for the body to return to its starting position and momentarily come to rest is identical to the forward journey:

The total time period $T$ for one complete oscillation is the sum of the time taken to reach the wall and the time taken to return:

A Conceptual Note: Is this Simple Harmonic Motion (SHM)? While the motion is strictly periodic (repeating every $1\text{ s}$), it is not SHM. In Simple Harmonic Motion, the restoring force must be directly proportional to the displacement from the mean position ($F=-kx$). In our scenario, the magnitude of the electric force ($F=qE$) remains absolutely constant regardless of where the body is on the plane. It is a beautiful example of periodic, non-harmonic motion!