Question

What is the electric potential energy of an arrangement of two charges, -20.51 μC and -13.37 μC, separated by 16.26 cm?

Answer 1

Given:

The charge is q1 = -13.37 micro Coulombs

The charge q2 = -20.51 micro Coulombs

The distance between them is 16.26 cm =0.1626 m

To find the electric potential energy.

Step-by-step explanation:

The potential energy can be calculated by the formula

`U=(kq1q2)/(r)`

Here, k is a constant whose value is

`k=9*10^9\text{ N m}^2\text{ /C}^2`

On substituting the values, the potential energy will be

`\begin{gathered} U=(9*10^9*13.37*10^(-6)*20.51*10^(-6))/(0.1626) \\ =\text{ 15.18 J} \end{gathered}`

The electric potential energy is 15.18 J