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

Answer:

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

Explanation:

Force due to q1 on q3

$F_{13} = \frac{kq_1q_3}{r^2}$

here we will have

$F_{13} = \frac{(9\times 10^9)(8 \times 10^{-6})(6 \times 10^{-6})}{0.13^2}$

$F_{13} = 25.56 N$

similarly force due to q2 on q3

$F_{23} = \frac{kq_2q_3}{r^2}$

here we will have

$F_{23} = \frac{(9\times 10^9)(10 \times 10^{-6})(6 \times 10^{-6})}{0.13^2}$

$F_{23} = 31.95 N$

Now net force along x direction will be given as

$F_x = F_{13}cos\theta + F_{23} cos\theta$

$F_x = (31.95 + 25.56) \times \frac{0.12}{0.13}$

$F_x = 53.1 N$

Now net force along y direction will be given as

$F_y = F_{13}sin\theta - F_{23} sin\theta$

$F_y = (31.95 - 25.56) \times \frac{0.05}{0.13}$

$F_y = 2.46 N$

Now net resultant force on q3

$F = \sqrt{F_x^2 + F_y^2}$

$F = 53.2 N$

direction of force is given by

$tan\theta = \frac{F_y}{F_x}$

$tan(\theta) = \frac{2.46}{53.1}$

$\theta = 2.65 degree$