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.

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

that would be b i just took the test : )