Question

A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction.part a the particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 1.60×10−18 j . in what direction and through what potential difference vb−va does the particle move?

Answer 1

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

$\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0$

where:

$\Delta K=+1.6\cdot 10^{-18} J$ is the change in kinetic energy of the particle

$q=3.2\cdot 10^{-19} C$ is the charge of the particle

$\Delta V =V_b-V_a$ is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

$\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V$