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

User Tuban
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1 Answer

10 votes
10 votes

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

User Melroy Van Den Berg
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2.7k points
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