Question

Two charges of -4.0 μC and -5.0 μC are held fixed at the lower two corners of an equilateral triangle of side length 2.0 m, in a vacuum. How much energy to move a point charge of +3.0 μC from a point midway between the other two charges is requi apex of the triangle?

Answer 1

`0.18225` Joules

Step-by-step explanation:

`L` = length of each side of triangle = 2 m

`r_(i)` = initial distance of charge from other two charges at the lower two corners = (0.5) L = (0.5) (2) = 1 m

`r_(f)` = final distance of point charge from other two charges at the apex of triangle = 2 m

`q` = point charge = 3 x 10⁻⁶ C

`q_(1)` = one of the charge on lower two corners = - 4 x 10⁻⁶ C

`q_(2)` = other charge on lower two corners = - 5 x 10⁻⁶ C

`E` = Energy required to move the point charge

Energy required to move the point charge

`E = (kqq_(1))/(r_(f)^(2)) + (kqq_(2))/(r_(f)^(2)) - (kqq_(1))/(r_(i)^(2)) - (kqq_(2))/(r_(i)^(2))`

`E = 0.18225` Joules