Question

Three identical point charges with charge q = +3.0 x 10-6 C are placed at each vertex of an equilateral triangle ABC as shown. If the side of the equilateral triangle is 0.01 m, find the resultant electric force on the charge at vertex B.

Answer 1

$$F_1=F_2=f=\frac{kq^2}{r^2}$$

$$f=\frac{(9\cdot10^9)(3\cdot10^{-6})^2}{0.01^2}=810\ N$$

$$F=\sqrt{f^2+f^2+2ff\cos{60}}=f\sqrt{2+2\cos{60}}\\F=810\sqrt{2+2\cos{60}}=1400\ N$$