Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The net force on the +1.0 nC charge is $1.2\times 10^{-5}~N$ directed upward.
By symmetry, the horizontal component of each force on the +1.0 nC charge cancels out. The net force on the +1.0 nC charge is the sum of the vertical component of each force. We can find the magnitude of the upward force on the +1.0 nC charge by the -6.0 nC charge. $F_u = \frac{(k)(1.0~nC)(6.0~nC)}{r^2}$ $F_u = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(6.0\times 10^{-9}~C)}{(0.050~m)^2}$ $F_u = 2.2\times 10^{-5}~N$ We can find the downward force on the +1.0 nC charge which is the sum of the vertical component from each of the two +2.0 nC charges. $F_d = 2\times \frac{(k)(1.0~nC)(2.0~nC)}{r^2}~cos(45^{\circ})$ $F_d = 2\times \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.050~m)^2}~cos(45^{\circ})$ $F_d = 1.0\times 10^{-5}~N$ We can find the net upward force on the +1.0 nC charge. $F = F_u-F_d$ $F = 2.2\times 10^{-5}~N - 1.0\times 10^{-5}~N$ $F = 1.2\times 10^{-5}~N$ The net force on the +1.0 nC charge is $1.2\times 10^{-5}~N$ directed upward.