Three charges lie on a line: A = +5.0 µC at x = 0, B = -2.0 µC at x = 0.20 m, and C = +3.0 µC at x…

Step 1. A on B. The distance is $0.20$ m. $F = \dfrac{kq_Aq_B}{r^2}$. The top is $9.0 \times 10^9 \times 5.0 \times 10^{-6} \times 2.0 \times 10^{-6}\ \text{N m}^2 = 0.090\ \text{N m}^2$ and the bottom is $(0.20\ \text{m})^2 = 0.040\ \text{m}^2$. So $F = 2.25\ \text{N}$. It is an attraction, towards A, so $-2.25$ N.

Step 2. C on B. The distance is $0.30$ m. $F = \dfrac{kq_Cq_B}{r^2}$. The top is $9.0 \times 10^9 \times 3.0 \times 10^{-6} \times 2.0 \times 10^{-6}\ \text{N m}^2 = 0.054\ \text{N m}^2$ and the bottom is $(0.30\ \text{m})^2 = 0.090\ \text{m}^2$. So $F = 0.60\ \text{N}$. It is an attraction, towards C, so $+0.60$ N.

Step 3. The net force on B. $-2.25\ \text{N} + 0.60\ \text{N} = -1.65\ \text{N}$. The size is $1.65$ N, towards A.

Step 4. The net force on C. A is $0.50$ m from C. For A on C, the top is $9.0 \times 10^9 \times 5.0 \times 10^{-6} \times 3.0 \times 10^{-6}\ \text{N m}^2 = 0.135\ \text{N m}^2$ and the bottom is $(0.50\ \text{m})^2 = 0.25\ \text{m}^2$. So $F_{AC} = 0.54\ \text{N}$, a push to $+x$. B is $0.30$ m from C. For B on C, the top is $0.054\ \text{N m}^2$ and the bottom is $0.090\ \text{m}^2$. So B pulls C towards B with $0.60\ \text{N}$, to $-x$. So $F = 0.54\ \text{N} - 0.60\ \text{N} = -0.06\ \text{N}$.