Answer:
[tex] 0.18225 [/tex] Joules
Explanation:
[tex]L[/tex] = length of each side of triangle = 2 m
[tex]r_{i}[/tex] = initial distance of charge from other two charges at the lower two corners = (0.5) L = (0.5) (2) = 1 m
[tex]r_{f}[/tex] = final distance of point charge from other two charges at the apex of triangle = 2 m
[tex]q[/tex] = point charge = 3 x 10⁻⁶ C
[tex]q_{1}[/tex] = one of the charge on lower two corners = - 4 x 10⁻⁶ C
[tex]q_{2}[/tex] = other charge on lower two corners = - 5 x 10⁻⁶ C
[tex]E[/tex] = Energy required to move the point charge
Energy required to move the point charge
[tex]E = \frac{kqq_{1}}{r_{f}^{2}} + \frac{kqq_{2}}{r_{f}^{2}} - \frac{kqq_{1}}{r_{i}^{2}} - \frac{kqq_{2}}{r_{i}^{2}}[/tex]
[tex]E = 0.18225 [/tex] Joules
