Find the x and ycomponents the magnitude and the direction o

Find the x- and y-components, the magnitude, and the direction of the electric field at x=0.150m, y=-0.400m

Exercise 21.49 n a rectangular coordinate system a positive point charge q 6.00-10-9C is placed at the point z +0 150 m, y = 0, and an identical point charge is placed at z.0 150 m, y = 0

Solution

point, P = (0.150, - 0.400) m

displacement vector from first charge,

r1 = (0.150, -0.400) - (0.150, 0) = - 0.40j

r1cap = r1 / |r1|

and E = ( k q / |r|^2) r_cap

hence E = k q r / |r|^3

E1 = (9 x 10^9 x 6 x 10^-9) (- 0.40 j) / (0.40^3)

E1 = - 337.5 j

due to another charge,

r2 = (0.150, -0.400) - (- 0.150, 0) = 0.30i - 0.40j

|r2| = sqrt(0.83^2 + 0.4^2) = 0.5 m

E2 = (9 x 10^9 x 6 x 10^-9 )(0.30i - 0.40j) / (0.5^3)

E2 = 129.6 i - 172.8 j N/C

E = E1 + E2 = 129.6i - 510.3j N/C

magniude = sqrt(129.6^2 + 510.3^2) = 526.5 N/C

direction = 360 - tan^-1(510.3 / 129.6) = 284.25 deg counter clock wise direction from +x axis.