**Electric Potential Energy of Three Point Charges**

**Solution Below:**

Part A

Three equal point charges, each with charge 1.30 μC, are placed at the vertices of an equilateral triangle whose sides are of length 0.500 m. What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Use ε

Use ε

_{0}= 8.85×10^{−12}C^{2}/(N*m^{2}) for the permittivity of free space.Mastering Physics says to use ε_{0} but we don’t have to (we can just use k directly). The formula to find the potential energy between two charges is:

U = (k*q_{1}*q_{2})/r

And since there are 3 pairs of charges (1-2, 1-3, and 2-3), we can solve for the potential energy of the system if we multiply the above by 3 (k is the Coulomb constant, which equal’s 8.99*10^{9}N*m^{2}/C^{2}). Note that each charge is 1.3 μC, which equals 1.3*10^{-6}C:

U = 3*(k*q_{1}*q_{2})/r

U = 3*((8.99*10^{9})*(1.3*10^{-6})*(1.3*10^{-6}))/0.5

U = 9.11*10^{-2} J

9.11*10^{-2} J

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the coulomb’s constant (k)suppose to be 8.99*10^9N*m2/C2, not 8.99*10-9N*m2/C2.

Thank you for the good catch!

cool!!! thanks

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