Question

FIGURE Q25.6 shows the electric potential along the x-axis.

a. Draw a graph of the potential energy of a $0.1\mathrm{C}$charged particle. Provide a numerical scale for both axes.

b. If the charged particle is shot toward the right from FIGURE Q25.6 $x=1\mathrm{m}$with$1.0\mathrm{J}$of kinetic energy, where is its turning point? Use your graph to explain.

Short answer

1. The potential energy graph looks like this

b. The motion's turning point is located at$x=3\mathrm{m}$

Step-by-step solution

Given Information

The charge of a particle is $q=0.1\mathrm{C}$

Graph of electric potential along the x axis

:Draw a graph of a 0.1C charged particle's potential energy.

(a)

When electric potential equals potential energy required, the graph of potential energy and electric potential are identical.

Therefore, the graph of potential energy is as follows:

Charge particle's turning point.

A turning point is a point at which the particles have lost all of their kinetic energy.

The particle is shot with kinetic energy of $1\mathrm{J}$at $x=1\mathrm{m}$, hence the total energy of the particle at $x=1\mathrm{m}$

$\begin{array}{rcl}K+U& =& 1+1\\ & =& 2\mathrm{J}\end{array}$

As there will be no kinetic energy, the turning point of the motion will be when the particle has a potential energy of $2\mathrm{J}$.

As can be seen from the graph, the particle's potential energy is $2\mathrm{J}$ at $x=3\mathrm{m}$, and so the motion's turning point is $x=3\mathrm{m}$.