Question

A negatively charged particle is moving in the positive x-direction when it enters a region with a uniform electric field pointing in the positive x-direction. Which graph gives its position as a function of time correctly? (its initial position is $x = 0$ at $t = 0$)
(A)

(B)

(C)

(D)

Answer 1

Hint: We can see that the acceleration of the particle will be a constant negative value since it is acted by a uniform force in the negative x-direction. We can calculate the slope in each of the graphs and see how the slope changes with time. The slope in displacement-time graphs gives the velocity of the particle. Finding the slope of a velocity-time graph gives us the acceleration which should be a negative constant.

Complete step by step answer:
The question states that negatively charged particles move in the positive x-direction. Thus it will have a velocity in the positive x-direction. Due to this initial velocity, it will move towards the positive x-direction due to which its x-component will increase.
Now it is given that there is an electric field present in the positive x-direction. This electric field is directed towards the positive x-direction which means that any positive charge will feel an electrostatic force towards the positive x-direction due to which the positive charge will start accelerating towards the positive x-direction.
However, the charge we have taken into consideration is a negatively charged particle. Thus on this negatively charged particle, due to the electric field present towards the positive x-direction, a net electrostatic force will act towards the negative x-direction. Due to this force, the negatively charged particle will have an acceleration towards the negative x-direction.
However, the particle already has a positive velocity directed towards the positive x-direction. So due to the negative acceleration, this velocity will start decreasing and will reverse in its direction after it reaches $0$. We will consider only the portion till where it remains in the positive x-axis.
In the meantime, the displacement will increase till the time the particle has a positive velocity. Then as the particle starts gaining negative velocity, the displacement starts decreasing.
This type of motion can only be seen in the third graph of the option. We know that the slope of the displacement time graph gives the velocity. Calculating the slope in the graph, we will see that the slope is initially positive, but decreases to zero and then becomes negative. This is similar to what we have analysed.
Therefore the correct answer is option (C).

Note:
Here we will consider that the particle is acted upon by a uniform force which does not change. This is true because it is already given that the electric field encountered by the particle is uniform in nature. However, we will also assume that this electric field is not time-varying either.