An alpha particle (m = 6.7 * 10-27 kg, q = +3.2 *10-19 C) has a speed of 20 km/s at point A and moves to point B where it momentarily stops.Only electric forces act on the particle during this motion.Determine the electric potential difference VA -VB.

Question

Quizplus · Accepted Answer

Since only electric forces act on the particle, the change in its kinetic energy is equal to the work done by the electric force:
ΔK = W = qΔV
where ΔK is the change in kinetic energy, q is the charge of the particle and ΔV is the potential difference.
At point A, the kinetic energy of the particle is K = (1/2)mv^2 = 4.472×10^-11 J
At point B, the kinetic energy of the particle is zero.
Therefore, ΔK = -4.472×10^-11 J
The charge of the particle is q = +3.2×10^-19 C
Substituting these values into the equation gives:
-4.472×10^-11 J = (3.2×10^-19 C)ΔV
Therefore, ΔV = (-4.472×10^-11 J) / (3.2×10^-19 C) = -139750 V ≈ -140 V
The electric potential difference VA - VB is VA - VB = -ΔV = 140 V
However, the question asks for the potential difference in the form VA - VB, where VA is the potential at point A and VB is the potential at point B. Since the particle moves from point A to point B, VA > VB and therefore, VA - VB = 140 V is negative.
Converting to the given options, we get (F1F1F1S1−F1F1F104.2 V) as the closest match. Therefore, the answer is B.

An Alpha Particle (M = 6