Three point charges are arranged the x axis Charge q1 410 m

Three point charges are arranged the x axis. Charge q_1 = -4.10 m C is located at x = 0.240 m and charge q_2 = 200 nC is at x = 0.350 m. A positive point charge q_2 is located at the origin. What must the value of q_3 b for the net force on this point charge to have magnitude 4.50 mu N? What is the direction of the net force on f_b Where along the x axis can q_3 be placed and the net force on it be zero, other than the initial answers of x = + infinity and x = - infinity?

Solution

Here ,

q1 = -4.1 nC

x1 = 0.240 m

q2 = 2 nC

x2 = 0.35 m

part A) F3 = 4.5 uN

Now , for the net force on charge q3

4.5 *10^-6 = 9 *10^9 * q3 * 10^-9 * (4.1/0.240 - 2/0.35)

solving for q3

q3 = 4.4 *10^-8 C

the charge q3 is 4.4 *10^-8 C

part B)

the direction of force on charge q3 is positive

part c)

for the force on q3 to be zero

point must be right of q2

k * (4.1)/(x - 0.240)^2 = k * 2/(x - 0.35)^2

(4.1)/(x - 0.240)^2 = 2/(x - 0.35)^2

solving for x

x = 0.605 m

the position of q3 can be 0.605 m