Question

A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential energy of the pair of charges is +5.4×10−8J+5.4×10−8J. When the second charge is moved to point bb, the electric force on the charge does −1.9×10−8J−1.9×10−8J of work.What is the electric potential energy of the pair of charges when the second charge is at point b?

Answer 1

Step-by-step explanation:

The given data is as follows.

`U_(a) = 5.4 * 10^(-8) J`

`W_{/text{a to b}} = -1.9 * 10^(-8) J`

Electric potential energy (
`U_(b)`) = ?

Formula to calculate electric potential energy is as follows.

`U_(b) = U_(a) - W_{/text{a to b}}`

=
`5.4 * 10^(-8) J - (-1.9 * 10^(-8) J)`

=
`7.3 * 10^(-8) J`

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is
`7.3 * 10^(-8) J`.