Two equal positive charges q are at the base angles of an eq

Two equal positive charges q are at the base angles of an equilateral triangle of sides a = 23 cm. a) What are the magnitude and direction of the electric field at point P due to the charges at the base of the triangle. Point P is located at the vertex angle or the triangle and q has a charge of 5.2 muC. b) What is the voltage at point P due to the two charges at the base angles of the equilateral triangle?

Solution

electric field is given by

E = kQ/r^2

electric field at P due to one charge = E = 9*10^9*5.2*10^(-6)/(0.23)^2 = 884688.1 N/C

electric field at P due to anther charge= E = 9*10^9*5.2*10^(-6)/(0.23)^2 = 884688.1 N/C

net electric field at point P = sqrt(E^2 + E^2 + 2*E^2*cos(60)) = sqrt(3)*884688.1 = 1516355.4 N/C

and direction is along +y axis.

electric potential is given by

P = kQ/R

potential at point P = 9*10^9*5.2*10^(-6)/(0.23) + 9*10^9*5.2*10^(-6)/(0.23) = 406956.52 volt