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 :

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.

Explanation:

The potential energy can be calculated by the formula

[tex]U=\frac{kq1q2}{r}[/tex]

Here, k is a constant whose value is

[tex]k=9\times10^9\text{ N m}^2\text{ /C}^2[/tex]

On substituting the values, the potential energy will be

[tex]\begin{gathered} U=\frac{9\times10^9\times13.37\times10^{-6}\times20.51\times10^{-6}}{0.1626} \\ =\text{ 15.18 J} \end{gathered}[/tex]

The electric potential energy is 15.18 J

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