Problem 27 - University Physics with Modern Physics

Question

A proton is traveling horizontally to the right at \(4.50 \times\) \(10^{6} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Step-by-Step Solution

Answer

For proton, the electric field is \(3.16 \times 10^4\ \mathrm{N/C}\) to the left. Time to stop is \(1.42 \times 10^{-9} \mathrm{s}\). For electron, field is the same magnitude, opposite direction.

Step 1: Analyze the Problem

We need to stop a proton over a distance of 3.20 cm using an electric field. Given the proton's initial velocity of \(4.50 \times 10^6 \ \mathrm{m/s}\), we need to find the electric field that will decelerate it to rest.

Step 2: Understand the Forces Involved

The force on the proton by the electric field \(F\) is given by \(F = qE\), where \(q\) is the charge of the proton \(1.60 \times 10^{-19}\ \mathrm{C}\), and \(E\) is the electric field strength.

Step 3: Calculate the Required Deceleration

Using the kinematic equation \(v_f^2 = v_i^2 + 2a x\), with \(v_f = 0\), \(v_i = 4.50 \times 10^6\ \mathrm{m/s}\), and \(x = 3.20 \ \mathrm{cm} = 0.032\ \mathrm{m}\), solve for \(a\). This gives us \(a = \frac{-(v_i^2)}{2x}\).

Step 4: Substitute and Solve for Electric Field

Using \(F = ma\) and \(F = qE\), substitute \(ma = qE\) to get \(E = \frac{ma}{q}\). Substitute the known values to find \(E\).

Step 5: Determine Direction of Electric Field

Since the proton is traveling to the right and needs to stop, the electric field must point to the left to decelerate the proton with a force opposite to its motion.

Step 6: Calculate Time to Stop

Use the equation \(v_f = v_i + at\). Solving for \(t\) gives \(t = \frac{-v_i}{a}\), using \(a\) calculated previously.

Step 7: Analyze the Scenario for an Electron

Repeat the calculation for an electron using its charge \(-1.60 \times 10^{-19}\ \mathrm{C}\). Determine that a field of the same magnitude as calculated for the proton is needed, but it should point in the opposite direction (to the right).

Step 8: Calculate Electric Field for Electron

Since the magnitudes are the same as for the proton, the calculations remain unchanged except for considering the direction opposite to that for the proton.

Key Concepts

Proton Motion

Protons are positively charged particles found in the nucleus of an atom. When a proton moves through an electric field, its motion can be affected significantly.

• Because protons have a positive charge, they experience force in the direction of the electric field. However, if the field is in the opposite direction of the motion, the proton will slow down.
• In the context of our exercise, we need an electric field to bring a moving proton to rest. This requires the field to exert a force opposite to the proton's initial direction of motion.
• The direction of this electric force will be crucial in determining how quickly the proton can stop.

By understanding proton motion in an electric field, we can manipulate conditions to either speed up or slow down the proton based on our needs.

Kinematics in Physics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It's crucial in calculating various parameters like velocity, acceleration, and displacement.To bring a proton to rest using an electric field, we utilize kinematic equations.

• The equation we use is: \( v_f^2 = v_i^2 + 2a x \)
• Here, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(x\) is the displacement. In our exercise, \(v_f\) becomes zero since the proton comes to a halt.
• Rearranging the equation helps us solve for acceleration, \(a = \frac{-v_i^2}{2x}\).

This acceleration is due to the force exerted by the electric field, and understanding this helps us relate the motion with the forces at play.

Electric Force

Electric force is the interaction between two charged objects. This force is what we use to manipulate the motion of charged particles like protons and electrons.

• For a proton, the electric force is calculated using the equation \( F = qE \). Here, \( q \) is the charge, and \( E \) is the electric field.
• In our scenario, this force is used to stop the proton by exerting it in the opposite direction of motion.
• We combine this with the mass \(m\) of the proton to determine the electric field required: \(E = \frac{ma}{q}\).

This concept is pivotal to controlling particle motion, such as stopping a proton over a given distance. The direction and magnitude of the electric field determine the effectiveness of this control.