Question

Point charges of + 3nC and - 3nC are placed at the corners A and of an equilateral triangle ABC each side of which is 0.3 m. What is the force acting on a + 2nc charge placed at C?
(6$\times {10}^{-7}$ N parallel to AB)

Answer 1

Let $F_1$ and $F_2$ exerted by the charge placed at A and B point of the equilateral triangle on C point as shown in the figure

$F_1=\frac{K{q}_A{q}_c}{r^2}=\frac{9\times {10}^9\times 3\times {10}^{-9}\times 2\times {10}^{-9}}{(.3{)}^2}=6\times {10}^{-7}$

$F_2=\frac{K{q}_B{q}_c}{r^2}=\frac{9\times {10}^9\times (-3)\times {10}^{-9}\times 2\times {10}^{-9}}{(.3{)}^2}=-6\times {10}^{-7}$

both $F_1$ and $F_2$ make angle of 120^0 between each other

resultant of $F_1$ and $F_2$ is F

$\Rightarrow F^2=F_1^2+F_2^2+F_1{F}_{2}cos\theta$

$\Rightarrow (6\times {10}^{-7}{)}^2+(-6\times {10}^{-7}{)}^2+2\times 6\times {10}^{-7}\times (-6)\times {10}^{-7}cos{120}^0$

$\Rightarrow (36+36-36)\times {10}^{-14}=36\times {10}^{-14}$

$\Rightarrow F=6\times {10}^{-7}N$