Question

A charge of 65 nC is located at a distance of 3.40 cm from a charge of −42 nC. What are the x− and y− components of the electric field at a point P that is directly above the 65 nC charge at a distance of 1.40 cm? Point P and the two charges are on the vertices of a right triangle.

Answer 1

q1 = -42 nC
q2 = 65 nC

Distance between q1 & Point p = sqrt(0.034^2 + 0.014^2) = 0.0368 m

X component of the Electric Field,
Ex = - k*q1/r^2 * cos(45) i^
Ex = - (8.9*10^9 * 42*10^-9 ) / (0.0368)^2 * cos(45) i^
Ex = - 1.95*10^5 i^ N/C

Y component of the Electric Field,
Ey = k*q2/r2^2 - k*q1/r^2 * sin(45) j^
Ey = [8.9*10^9 * 10^-9] * [65/0.014^2 - 42/0.0368^2 * sin(45)] j^
Ey = 2.76*10^6 j^ N/C