In a rectangular coordinate system a positive point charge q = 6.00 x 10^-9C is placed at the point x = +0.150 m, y = 0, and an identical point charge is placed at x = -0.150 m, y = 0. (a) Find the x- and y-components, magnitude and the direction of the electric field at the following points (a) the origin; (b) x = 0.300 m, y = 0; (c) x = 0.150 m, y = -0.40; (d) x = 0, y=0.200 m (a) 0 N/C (b) 2667 N/C (c) 526 N/C , 284 deg counter clockwise from positive x axis

a) 0N/C

b) 2660.7N/C

c) 543.7N/C

Explanation:

To find the electric field in all these cases you take into account the x and y component of the total Electric field for each point.

a)

for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:

  ( 1 )

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r1: distance to the first charge = 0.150m

r:2 distance to the second charge = 0.150m

Due to in this case the distance r1=r2: you obtain, by replacing in (1):

b)

for P(0.300 , 0 ) you also have only the x component of E, and the electric field generated by each charge are directed toward right:

r1 = 0.300m+0.150m = 0.450m

r2 = 0.300m-0.150m = 0.150m

By replacing r1 and r2 in ( 1 ) you obtain:

c)

for P(0.150 , -0.40) you have both x and y components for E:

the second charge does not contribute for the x component of E.

To find r1 you use Pitagora's theorem:

r2 = 0.40m

the angle is obtain by using a simple trigonometric relation:

Then, by replacing the values of r1, r1, q, theta and k you obtain:

hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.