Question

A particle with a charge of -2.7 ?C and a mass of 3.8 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 36 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.

Answer 1

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:

$\frac{1}{2}mv^2 = q\Delta V$

where

$m=3.8\cdot 10^{-6} kg$ is the mass of the particle

$v=36 m/s$ is the final speed of the particle

q = -2.7 C is the charge

$\Delta V$ is the potential difference between the two points

Solving for $\Delta V$, we find

$\Delta V= \frac{mv^2}{q}=\frac{(3.8 \cdot 10^{-6} kg)(36 m/s)^2}{-2.7 C}=-0.0018 V$

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

$V_B - V_A = 0.0018 V$