A charge −1.3 × 10−5 C is fixed on the x-axis at 7 m, and a charge 1 × 10−5 C is fixed on the y-axis at 4 m. Calculate the magnitude of the resultant electric field E~ at the origin. Answer in units of N/C.

6104 N/C.

Explanation:

Given:

k = 8.99 × 10^9 Nm2/C^2

Qx = 1.3 × 10^-5 C

rx = 7 m

Qy = 1 × 10−5 C

ry = 4 m

E = F/Q

= kQ/r^2

Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2

= 2385.1 N/C.

Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2

= 5618.75 N/C

Eo = sqrt(Ex^2 + Ey^2)

= sqrt(3.157 × 10^7 + 5.69 × 10^6)

= 6104 N/C.

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.