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Georgia [21]
3 years ago
12

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.
Physics
1 answer:
katovenus [111]3 years ago
6 0

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

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