Question

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in electrical potential energy of the particle?

Answer 1

`7.2\cdot 10^(-19) J`

Step-by-step explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

`\Delta U = q \Delta V`

where

q is the magnitude of the charge of the particle

`\Delta V` is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

`q=3e=3\cdot 1.6\cdot 10^(-19) J=4.8\cdot 10^(-19)J`

- the potential difference is

`\Delta V=4.50 V`

So, the change in electrical potential energy is

`\Delta U=(1.6\cdot 10^(-19)C)(4.50 V)=7.2\cdot 10^(-19) J`