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Protons are positively charged particles found in an atom's nucleus. Observing their motion is crucial for understanding various physical phenomena. When a proton moves through an electric field, the field can exert a force on it, causing the proton to accelerate or decelerate. In our case, we need to stop the proton, initially traveling at a speed of \(4.50 \times 10^6 \text{ m/s}\). To accomplish this, an electric field must be applied in the opposite direction of the proton's motion. By calculating the force required to bring the proton to a stop, we can determine the properties of this electric field. The stopping distance we consider here is \(0.0320 \text{ m}\), converted from \(3.20\) cm. It is also significant to note that the charge of a proton is \(1.60 \times 10^{-19} \text{ C}\) and its mass is \(1.67 \times 10^{-27} \text{ kg}\). Understanding these values assists in determining how the proton will behave under the influence of an electric field.

Kinematics is the branch of mechanics dealing with motion without considering the forces that cause the motion. In this exercise, we use kinematic equations to find the necessary acceleration to bring the proton to rest. The key equation we use is:\[ v_f^2 = v_i^2 + 2ad \] where \( v_f \) is the final velocity (0 m/s when the proton stops), \( v_i \) is the initial velocity, and \( d \) is the stopping distance. By substituting the known values into the equation, we solve for the acceleration \( a \). This calculated acceleration is fundamental in determining how the electric field influences the proton, providing a direct link between motion and electric forces.

• Initial velocity \( v_i = 4.50 \times 10^6 \text{ m/s} \)
• Stopping distance \( d = 0.0320 \text{ m} \)
• Resulting acceleration \( a = -3.16 \times 10^{14} \text{ m/s}^2 \)

Recognize that the negative sign of acceleration indicates the proton is slowing down.

Forces are fundamental to understanding motion within physics. When a proton travels through an electric field, it experiences an electric force given by \( F = qE \), where \( q \) is the charge and \( E \) is the electric field strength. Newton's second law, \( F = ma \), helps us connect the force experienced by the proton to its mass and acceleration. For the proton, we calculate:\[ F = ma = (1.67 \times 10^{-27} \text{ kg}) \times (-3.16 \times 10^{14} \text{ m/s}^2) \]

• Force experienced by proton: \( F = -5.28 \times 10^{-13} \text{ N} \)
• Electric field strength required: \( E = \frac{F}{q} = -3.30 \times 10^{6} \text{ N/C} \)

The calculated electric field must be directed to the left to oppose the proton's motion, aligning with the required deceleration.

Electron dynamics under an electric field differ from protons due to smaller mass and negative charge. To stop an electron under the same conditions as a proton, we need to account for its different mass \(9.11 \times 10^{-31} \text{ kg}\) and charge \(-1.60 \times 10^{-19} \text{ C}\). While the electron experiences the same magnitude of acceleration, due to its lighter mass, the forces involved differ. For the electron:

• Acceleration \( a = -3.16 \times 10^{14} \text{ m/s}^2 \) remains the same.
• Experienced force \( F = (9.11 \times 10^{-31} \text{ kg}) \times (-3.16 \times 10^{14} \text{ m/s}^2) \approx -2.88 \times 10^{-16} \text{ N} \)
• Electric field needed: \( E = \frac{F}{-q} = 1.80 \times 10^{3} \text{ N/C} \)

The positive value of \( E \) indicates that the field should be directed to the right, the same as the electron's initial direction but opposite in sign to facilitate stopping.