Question

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 4.80×10−19 JJ . In what direction and through what potential difference Vb−VaVb−Va does the particle move?

Answer 1

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$

Answer 2

The potential difference between Vb−Va is 1 J. The work-energy that is needed to move a charged particle from one point to another in an electric field.

What is Electric potential?

The work-energy that is needed to move a charged particle from one point to another in an electric field.

The potential energy is completely converted to kinetic energy when a particle moves inside an electrical field,

$\Delta U = q \Delta V$

Where,

$\Delta U$- Change in Potential energy =  $4.8 \times 10^{-19}$  J

$q$ - Charge = $4.8 \times 10^{-19}$   C.

$\Delta V$ - change in Votage =  ?

Put the values in the formula,

$\Delta V = \dfrac {4.8 \times 10^{-19}}{4.8 \times 10^{-19}}\\\\\Delta V = 1 \rm \ J$

Therefore, the potential difference between Vb−Va is 1 J.

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