Question

The three charges in the figure below are at the vertices of an isosceles triangle. Let q = 3.00 nC and calculate the electric potential at the midpoint of the base. (Let d1 = 3.00 cm and d2 = 7.00 cm.)

Answer 1

electric potential = k * Q / r

distance of the bottom two charges from the middle of the base is 3 / 2 cm

so,

electric potential due to bottom two charges = 2 * 9 * 10^9 * 3 * 10^-9 / (1.5 * 10^-2)

distance of top charge from midpoint of base = sqrt(0.07^2 - (1.5 * 10^-2)^2)

electric potential due to upper charge = 9 * 10^9 * 3 * 10^-9 / sqrt(0.07^2 - (1.5 * 10^-2)^2)

net electric potential at the midpoint of the base = 2 * 9 * 10^9 * 3 * 10^-9 / (1.5 * 10^-2) + 9 * 10^9 * 3 * 10^-9 / sqrt(0.07^2 - (1.5 * 10^-2)^2)

net electric potential at the midpoint of the base = 3994.88711 V or 3.99488711 kV