Charges of +9.0 µC and +1.0 µC are fixed 0.40 m apart…

Part (a). Plan. The push of each fixed charge on the small charge must be equal and opposite.

Step 1. The position. $\dfrac{x}{d - x} = \sqrt{\dfrac{9.0}{1.0}} = 3$. So $x = 3(0.40\ \text{m} - x)$, which gives $4x = 1.20\ \text{m}$ and $x = 0.30$ m. That is $30$ cm from the $9.0\ \mu\text{C}$ charge, and $10$ cm from the $1.0\ \mu\text{C}$ charge.

Part (b). Step 2. The sign. The $1.0\ \mu\text{C}$ charge is pushed away from the $9.0\ \mu\text{C}$ charge. To keep it still, the middle charge $q_3$ must pull it back. So $q_3$ is negative.

Step 3. The size. For the $1.0\ \mu\text{C}$ charge, the pull of $q_3$ at $0.10$ m equals the push of the $9.0\ \mu\text{C}$ charge at $0.40$ m. The constant $k$ and the $1.0\ \mu\text{C}$ charge cancel, so $\dfrac{q_3}{(0.10\ \text{m})^2} = \dfrac{9.0\ \mu\text{C}}{(0.40\ \text{m})^2}$. Then $q_3 = 9.0\ \mu\text{C} \times \left(\dfrac{0.10}{0.40}\right)^2 = 0.5625\ \mu\text{C}$.