Aparticle 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?

The potential difference through which the particle moves is 5 V, in negative x-direction.

The given parameters;

The change in the potential energy of the particle is calculated by applying the principle of conservation of energy;

Thus, the potential difference through which the particle moves is 5 V, in negative x-direction.

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-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:

where:

is the change in kinetic energy of the particle

is the charge of the particle

is the potential difference

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

none, its a machine

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