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

Explanation:

The given data is as follows.

            $U_{a} = 5.4 \times 10^{-8} J$

        $W_{/text{a to b}} = -1.9 \times 10^{-8} J$

        Electric potential energy ($U_{b}$) = ?

Formula to calculate electric potential energy is as follows.

            $U_{b} = U_{a} - W_{/text{a to b}}$

                        = $5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)$

                        = $7.3 \times 10^{-8} J$

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is $7.3 \times 10^{-8} J$.