Answer:
0.0018 V
Explanation:
According to the law of conservation of energy, the kinetic energy gained by the particle is equal to the electric potential energy lost:
[tex]\frac{1}{2}mv^2 = q\Delta V[/tex]
where
[tex]m=3.8\cdot 10^{-6} kg[/tex] is the mass of the particle
[tex]v=36 m/s[/tex] is the final speed of the particle
q = -2.7 C is the charge
[tex]\Delta V[/tex] is the potential difference between the two points
Solving for [tex]\Delta V[/tex], we find
[tex]\Delta V= \frac{mv^2}{q}=\frac{(3.8 \cdot 10^{-6} kg)(36 m/s)^2}{-2.7 C}=-0.0018 V[/tex]
The particle has been accelerated by this potential difference: since it is a negative charge, it means that the particle has moved from a point at lower potential towards a point of higher potential.
So, since the initial point is A and the final point is B, the result is
[tex]V_B - V_A = 0.0018 V[/tex]
