Four equal charges of +2.5 µC are at the corners of a square of side 0.15 m…

Step 1. One neighbour. $F = \dfrac{kq^2}{a^2} = \dfrac{9.0 \times 10^9 \times (2.5 \times 10^{-6})^2}{(0.15)^2}\ \text{N} = 2.50\ \text{N}$.

Step 2. The opposite corner. The distance is $\sqrt{2}\,a$, so the distance squared is $2a^2$. $F' = \dfrac{F}{2} = 1.25\ \text{N}$, along the diagonal, away from the opposite corner.

Step 3. The two neighbours. They are at a right angle, with $2.50$ N each. Their sum is $\sqrt{2} \times 2.50\ \text{N} = 3.54\ \text{N}$, along the diagonal.

Step 4. The net force. The sum of the neighbours and the push of the opposite corner both lie along the diagonal, so we add them: $3.54\ \text{N} + 1.25\ \text{N} = 4.79\ \text{N}$, directed away from the centre.

Step 5. The charge at the centre. It must attract with $4.79$ N. The distance to a corner is $\dfrac{0.15}{\sqrt{2}} = 0.106$ m. From $F = \dfrac{k|Q|q}{r^2}$ we want $|Q|$, so $|Q| = \dfrac{Fr^2}{kq}$. The top is $4.79\ \text{N} \times (0.106\ \text{m})^2 = 0.0538\ \text{N m}^2$. The bottom is $9.0 \times 10^9\ \text{N m}^2/\text{C}^2 \times 2.5 \times 10^{-6}\ \text{C} = 2.25 \times 10^4\ \text{N m}^2/\text{C}$. So $|Q| = 2.39 \times 10^{-6}\ \text{C}$. It must attract, so it is negative.