Question

A particle with charge 1.60×10−19 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.The particle, initially at rest is acted on by the electricforce and moves from point a to point b along the x axis increasingits kinetic energy by 8.00 x 10 -19J In what direction and through what potential difference does the particlemove

If the particle moves from point b to point c in the ydirection what is the change in its potential energy

Answer 1

Answer:

Explanation:

According to  energy conservation principle;

$\delta \ K.E = \delta \ P.E$

$8.00*10^{-19} \ J = (1.6*10^{-19} \ C ) |V_b-V_a|$

$|V_b-V_a|= \frac {8.00*10^{-19} }{(1.6*10^{-19} ) }$

$|V_b-V_a|= 5 \ V$

The potential is positive hence the electric fielf=d is negative along the x-axis.

We can then say that the movement of the particle goes to the left through a potential difference of  $|V_b-V_a|= 5 \ V$.

There will be no significant change in the y-direction of the potential energy when the particle moves from point b to point c in the y-direction.