Two charges and are separated by 10cm on the x- axis. Q2 is at the origin and Q1 is 12cm to the right. a. What is the net Electric Force (magnitude and direction) on Q2 due to Q1?
b. Calculate the magnitude and direction of the net electric field -2.0 cm to the left of Q2.

Explanation:

Q₂ is at the origin and Q₁ is at 12 cm on x -axis

Force between Q₁ and Q₂

= K x Q₁ x Q₂ / .12 where k is a constant equal to 9x 10⁹

Force F = 9 x 10⁹ x Q₁ x Q₂ / .12²

= 625 x 10⁹ x Q₁Q₂ N

This is the force each one exerts on other . So electric force on Q₂ due to Q₁ is 625 x 10⁹ x Q₁Q₂ N . Its direction is towards left as both will repel each other.

b)

Electric field due to charge Q₂ at - 2 cm

= 9 x 10⁹ x Q₂ / .02²

= 22500 x 10⁹ x Q₂ N/C , towards left .

Electric field due to charge Q₁ at - 2 cm

= 9 x 10⁹ x Q₁ / .14²

= 459.18 x 10⁹ x Q₁  N/C , towards left

As both the fields are to the same direction

Net field = 75 x 10⁹ x Q₂ + 459.18 x 10⁹ x Q₁

10⁹ ( 22500 x Q₂ + 459.18  x Q₁ ) N/C .

Direction will be towards left.

a and b sometimes have a positive value