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?

We have three charges along the x-axis:

• $q_1 = +15.0 \times 10^{-6} \text{ C}$ at $x_1 = 2.00 \text{ m}$
• $q_2 = +6.00 \times 10^{-6} \text{ C}$ at $x_2 = 0 \text{ m}$
• $q_3$ at an unknown position $x$.

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, $0 < x < 2.00 \text{ m}$.

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, $F = k \frac{|q_a q_b|}{r^2}$:

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.

The X coordinate of $q_3$ is $\boxed{\text{0.775 m}}$.

.3:

Each sphere is in equilibrium under three forces:

• Tension ($T$) in the string, acting along the string.
• Gravitational force ($mg$), acting vertically downwards