Three point charges of magnitude 5 , 0 . 16 and 0 . 3 are located...

Using Coulomb's law the electrostatic force on& $A$~due to& $C$

${F}_{AC} = \dfrac{K{q}_{A}{q}_{C}}{\left({r}_{AC}\right)2} = \dfrac{k \times 5 \times 0 . 3 \times {10}^{- 12}}{9 \times {10}^{- 4}}= \dfrac{9 \times {10}^{9} \times 5 \times 0 . 3 \times {10}^{- 12}}{9 \times {10}^{- 4}}$

$= 1 . 5 \times 10 = 15N$

Similarly, the force on~$A$~due to& $B$

${F}_{AB} = \dfrac{K{q}_{A}{q}_{B}}{\left({r}_{AB}\right)2} = \dfrac{9 \times {10}^{9} \times 5 \times 0 . 16 \times {10}^{- 12}}{9 \times {10}^{- 4}} = 8N$

We can see that the forces are perpendicular to each other. Therefore, the net force experienced by charge at $A$~is,

$F = \sqrt{{FAC}^{2} + {FAB}^{2}}= \sqrt{{15}^{2} + {8}^{2}} = \sqrt{289} = 17N$