The net force on the second charge is (665.15i - 4,415.87 j) N.
The given parameters:
- Charge, q2 = 14.33 µC
- Charge q3 = -17.18 µC
- Charge q1 = 5.15 µC
Distance between the first, second and third charge
The distance between the first, second and third charge is calculated as follows;
[tex]r_{12} = \sqrt{(1--2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{10} = 3.16 \ cm\\\\r_{23} = \sqrt{(1-0)^2 + (1--1)^2} = \sqrt{1^2 + 2^2} = \sqrt{5} = 2.24 \ cm[/tex]
The force on the second charge due to the first charge is calculated as follows;
[tex]F_{12} = \frac{kq_1 q_2}{r^2} \\\\F_{12} = \frac{9\times 10^9 \times 5.15 \times 10^{-6} \times 14.33 \times 10^{-6} }{(0.0316)^2} (i)\\\\F_{12} = 665.15i \ N\\\\[/tex]
The force on the second charge due to the third charge is calculated as follows;
[tex]F_{23} = \frac{kq_2q_3}{r_{23}^2} \\\\F_{23} = \frac{9\times 10^9 \times 14.33 \times 10^{-6} \times (-17.18 \times 10^{-6} )}{(0.0224)^2} (j)\\\\F_{23} = -4,415.87 j \ N[/tex]
The net force on the second charge is calculated as follows;
[tex]F_2_{net} = (665.15 i - 4,415.87j)N[/tex]
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