Force due to -8 uC charge is given as
[tex]F_1 = \frac{kq_1q}{r^2}[/tex]
[tex]F_1 = \frac{(9\times 10^9)(8\times 10^{-6})(6 \times 10^{-6})}{0.3^2}[/tex]
[tex]F_1 = 4.8 N[/tex]
Force due to -5 uC charge is given as
[tex]F_2 = \frac{kq_2q}{r^2}[/tex]
[tex]F_2 = \frac{(9\times 10^9)(5\times 10^{-6})(6 \times 10^{-6})}{0.3^2}[/tex]
[tex]F_2 = 3 N[/tex]
Now we know that two forces on 6 uC charge is perpendicular to each other
so we will have
[tex]F = \sqrt{F_1^2 + F_2^2}[/tex]
[tex]F = \sqrt{3^2 + 4.8^2}[/tex]
[tex]F = 5.66 N[/tex]
so net force due to both charges on 6 uC is 5.66 N
