Question

Consider three point charges at the corners of a triangle, as shown in the figure, where q1 = 8.00 x10^-9C, q2 = −4.00 × 10^−9 C, and q3 = 6.00 × 10^−9 C. Find the magnitude and direction of the resultant force on q3

Answer 1

$F_{23} = \frac{kq_2q_3}{4^2} = \frac{9 \times 10^9 \times 4 \times 6 \times 10^{-18}}{16} \\ F_{23} = -13.5 \times 10^{-9} \;N \\ F_{23x}= -13.5 \times 10^{-9} \;N \\ F_{23y}= 0 \\ F_{13} = \frac{kq_1q_2 }{5^2} = \frac{9 \times 10^9 \times 8 \times 6 \times 10^{-18}}{25} \\ F_{13} = 17.28 \times 10^{-9} \;N \\ F_{13x} = F_{13} \times cos(37) = 17.28 \times 10^{-9} \times 0.7986 = 13.799 \times 10^{-9} \;N \\ F_{13y} = F_{13} \times sin (37) = 17.28 \times 10^{-9} \times 0.6018 = 10.399 \times 10^{-9} \;N$

Total force:

$F_x=F_{23x}+F_{13x}=-13.5 \times 10^{-9} + 13.799 \times 10^{-9} = 0.299 \times 10^{-9} \; N \\ F_y=F_{23y}+F_{13y}= 0 + 10.399 \times 10^{-9} = 10.3399 \times 10^{-9} \; N \\ Magnitude= \sqrt{F_x^2+F_y^2}= 1.0344 \times 10^{-8} \;N$

angle with +ve x axis $=arctan(\frac{F_y}{F_x})= 72.07 \; degrees$