A point charge -2.06nC is at the point x = 0.60 m, y = 0.80 m , and a second poi

Question

A point charge -2.06nC is at the point  x = 0.60 m, y = 0.80 m , and a second point charge  5.57nC is at the point  x = 0.60 m , y = 0.

What is the magnitude of the electric field set up at the origin from q1?

What is the magnitude of the electric field set up at the origin from q2?

What is the x component of the net electric field at the origin?

What is the y component of the net electric field at the origin?

Calculate the magnitude of the net electric field at the origin due to these two point charges.

Calculate the direction of the net electric field at the origin due to these two point charges.

Explanation / Answer

1. x = 0.6 m, y = 0.8 m, r = 1.0 m

E1 = kq/r2 = (9 x 109 N m2/C2)(-2.06 nC / (109 nC/C) ) / (1.0 m)2 = 18.54 N/C

2. x = 0.6 m, y = 0, r = 0.6 m

E2 = kq/r2 = (9 x 109 N m2/C2)(5.57 nC / (109 nC/C) ) / (0.6 m)2 = 139.75 N/C

3. Find what the angle to the x-axis is for the first point charge.

cos (theta) = 0.6 m / 1 m

theta = 53.13o

Ex = (18.54 N/C)(cos 53.13o) - 139.75 N/C = -128.626 N/C

4. only the first point charge affects Ey

Ey = (18.54 N/C)(sin 53.13o) = 14.832 N/C

5. Pythagorean theorem

Enet2 = Ex2 +Ey2

Enet2 = (-128.626 N/C)2 + (14.832 N/C)2

Enet = 129.478 N/C

6. find the angle using trig

tan (theta) = Ey/Ex = (14.832 N/C) / (-128.626 N/C)

theta = 173.422o

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