Question

A 8 µC charge q1 and a 10 µC charge q2 are 0.05 m from the x-axis. A 6 µC charge q3 is 0.12 m from the y-axis. The distances d13 and d23 are 0.13 m. Find the magnitude and direction of the resulting vector R. Round to one decimal place.

Answer 1

The magnitude is 53N and the direction is 3 degrees. I hope this helps!

Answer 2

Net resultant force is 53.2 N at an angle of 2.65 degree from horizontal

Step-by-step explanation:

Force due to q1 on q3

`F_(13) = (kq_1q_3)/(r^2)`

here we will have

`F_(13) = ((9* 10^9)(8 * 10^(-6))(6 * 10^(-6)))/(0.13^2)`

`F_(13) = 25.56 N`

similarly force due to q2 on q3

`F_(23) = (kq_2q_3)/(r^2)`

here we will have

`F_(23) = ((9* 10^9)(10 * 10^(-6))(6 * 10^(-6)))/(0.13^2)`

`F_(23) = 31.95 N`

Now net force along x direction will be given as

`F_x = F_(13)cos\theta + F_(23) cos\theta`

`F_x = (31.95 + 25.56) * (0.12)/(0.13)`

`F_x = 53.1 N`

Now net force along y direction will be given as

`F_y = F_(13)sin\theta - F_(23) sin\theta`

`F_y = (31.95 - 25.56) * (0.05)/(0.13)`

`F_y = 2.46 N`

Now net resultant force on q3

`F = √(F_x^2 + F_y^2)`

`F = 53.2 N`

direction of force is given by

`tan\theta = (F_y)/(F_x)`

`tan(\theta) = (2.46)/(53.1)`

`\theta = 2.65 degree`