This proton has been “fired” from the left towards the right, in the uniform ele

Question

This proton has been “fired” from the left towards the right, in the uniform electric field shown. It will come to a stop, and then move back to the left.

Initially, its acceleration is:
[] to the right
[] to the left
[] zero

At the instant it comes to a stop, its acceleration is:
[] to the right
[] to the left
[] zero

After it comes to a stop, its acceleration is:
[] to the right
[] to the left
[] zero

During this process, the magnitude of the force on the proton is:
[] strong when it's on the left, and weak when it's on the right
[] weak when it's on the left, and strong when it's on the right
[] a constant strength throughout the process.

Explanation / Answer

1. Initially, the acceleration is due to the electric field. The electric field is towards left so force applied on the proton is towards left. Hence, acceleration is towards left. This is why the velocity of the proton moving towards right is going to decrease until it stops completely and then starts to move towards left. This can be understood in a different way like this: a = dv/dt. Since due to electric field in opposite direction, the velocity of proton is going to decrease, so a = dv/dt = negative = towards left if the particle is moving right.

2. At the instant it comes to a stop, the velocity is zero. a = dv/dt = 0. So acceleration is zero.

3. After it comes to a stop, there is going to be an acceleration towards left again because the motion is going to be towwards left due to the electric field. a = dv/dt = positive = towards left as the particle is also moving towards left.

4. Force is constant throught out the process. F = qE where q=total charge of the proton

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