Energy = PE = PE₁₂+PE₁₃+PE₁₄+PE₂₃+PE₂₄+PE₃₄=
=kq₁q₂/a +kq₁q₃/a+kq₁q₄/a+ kq₂q₃/a+kq₂q₄/a +kq₃q₄/a .
Since q₁=q₂=q₃=q₄=q=1C, and a=1 m
PE=6q²/a=6•1²/1= 6 J
Four charges are arranged at the corners of a regular tetrahedron (a pyramid with four sides that are identical equilateral triangles). The charges are 1 Coulomb each, and the sides of the tetrahedron are all 1 meter. How much energy in Joules is required to assemble this arrangement of charge?
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