Three charges are placed along the x-axis. q1 = +15.0 μC is at x = 2.00 m, and q2 = +6.00 μC is at x = 0 m. Where must a third charge, q3, be placed so that the net force on it is zero?

Alright sucess — let's do this. Exercise 1.2: Step 1: Identify the charges and their positions. We have three charges along the x-axis: q_1 = +15.0 × 10^-6 C at x_1 = 2.00 m q_2 = +6.00 × 10^-6 C at x_2 = 0 m q_3 at an unknown position x. Step 2: Determine the region where the net force on q_3 can be zero. Since q_1 and q_2 are both positive, for the forces on q_3 to cancel, q_3 must be located between q_1 and q_2. If q_3 is positive, q_1 and q_2 will both repel it, causing forces in opposite directions. If q_3 is negative, q_1 and q_2 will both attract it, also causing forces in opposite directions. Therefore, 0 < x < 2.00 m. Step 3: Apply Coulomb's Law for equilibrium. For the net force on q_3 to be zero, the magnitude of the force exerted by q_1 on q_3 (F_13) must be equal to the magnitude of the force exerted by q_2 on q_3 (F_23). The distance between q_2 and q_3 is r_23 = x. The distance between q_1 and q_3 is r_13 = 2.00 - x. Using Coulomb's Law, F = k (|q_a q_b|)/(r^2): F_13 = F_23 k (|q_1 q_3|)/(r_13)^2 = k (|q_2 q_3|)/(r_23)^2 (|q_1|)/((2.00 - x)^2) = (|q_2|)/(x^2) Step 4: Substitute values and solve for x. 15.0 × 10^-6 C(2.00 - x)^2 = 6.00 × 10^-6 Cx^2 (15.0)/(6.00) = ((2.00 - x)^2)/(x^2) 2.5 = ((2.00 - x)/(x))^2 Take the square root of both sides. Since 0 < x < 2.00, both x and 2.00-x are positive, so we take the positive root. sqrt(2.5) = (2.00 - x)/(x) 1.5811 = (2.00 - x)/(x) 1.5811x = 2.00 - x 1.5811x + x = 2.00 2.5811x = 2.00 x = (2.00)/(2.5811) x ≈ 0.7748 m The X coordinate of q_3 is 0.775 m. --- Exercise 1.3: Step 1: Draw a free-body diagram for one sphere. Each sphere is in equilibrium under three forces: Tension (T) in the string, acting along the string. Gravitational force (mg), acting vertically downwards