Question

A proton is placed midway between points \(A\) and \(B .\) The potential at point \(A\) is \(-20 \mathrm{~V},\) and the potential at point \(B+20 \mathrm{~V}\). The potential at the midpoint is \(0 \mathrm{~V}\). The proton will a) remain at rest. b) move toward point \(B\) with constant velocity. c) accelerate toward point \(A\). d) accelerate toward point \(B\). e) move toward point \(A\) with constant velocity.

Short answer

Answer: The proton will c) accelerate toward point A.

Step-by-step solution

Understand the given quantities

We are given the following information: - The potential at point A: \(-20V\) - The potential at point B: \(+20V\) - The potential at the midpoint: \(0V\) - The proton is placed at the midpoint.

Relate electric potential to electric field

We know the relationship between electric potential and electric field is given by: \(E = -\frac{\Delta V}{\Delta x}\) Where: \(E\) is the electric field, \(\Delta V\) is the change in electric potential, and \(\Delta x\) is the change in position along the direction of the electric field.

Calculate the electric field

In this case, we want to calculate the electric field between the midpoint, where the proton is placed, and either point A or point B. Since the system is symmetrical, the magnitude of the electric field will be the same in both cases. Let's consider the midpoint and point B: \(\Delta V_{B} = 20 V - 0 V = 20 V\) The proton is placed midway, so let's assume \(\Delta x = d\). \(E = -\frac{20V}{d}\) The negative sign indicates that the direction of the electric field is from point B to the midpoint.

Determine the force on the proton

Now, let's calculate the electric force experienced by the proton due to the electric field: \(F = qE\) Where: \(F\) is the force experienced by the proton, \(q\) is the charge of the proton, and \(E\) is the electric field between the proton and point B. As we know, the charge of the proton is positive, \(q = +1.6\times10^{-19}C\). Therefore, the force experienced by the proton will be: \(F = (1.6\times10^{-19}C)(-\frac{20V}{d})\) \(F = -\frac{32\times10^{-19}V}{d}\) The negative sign indicates that the force is directed towards point A, since it is opposite to the direction of the electric field.

Analyze the motion of the proton

As the calculated force is non-zero and directed towards point A, the proton will experience an acceleration towards point A due to Newton's second law: \(F = ma\) Where: \(F\) is the net force experienced by the proton, \(m\) is the mass of the proton, and \(a\) is the acceleration experienced by the proton. Since the force experienced by the proton is acting towards point A, the proton will accelerate towards point A. Therefore, the correct answer is: c) accelerate toward point A.