Answer:
[tex]\vec F_{A} = -67500\,N\cdot (i + j)[/tex]
Explanation:
The position of each point are the following:
[tex]A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)[/tex]
Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:
[tex]\vec F_{A} = \vec F_{AB} + \vec F_{AC}[/tex]
[tex]\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j[/tex]
[tex]\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)[/tex]
[tex]\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)[/tex]
[tex]\vec F_{A} = -67500\,N\cdot (i + j)[/tex]
