Question

What is the magnitude of the force F on the -10 nC charge in (Figure 1)? Suppose that a = 1.8 cm. Part B What is the direction of the force F on the - 10 nC charge in the figure? Give your answer as an angle measured cw or cw (specify which) from the + x - axis.

Answer 1

From the figure angle $\theta$ is given by

$\theta = \tan^{-1}\left ( \frac{AD}{CD} \right ) = \tan^{-1}\left ( \frac{1.8 cm}{3.0 cm} \right ) = 30.96^{0}$

Also

$AC = \sqrt{(3.0)^{2}+(1.8)^{2}} = 3.5 cm$

So Force on -10 nC charge due to 5.0 nC charge

$F_{BC} = \frac{kq_{B}q_{C}}{BC^{2}}= \frac{9\times 10^{9}\times 5\times 10^{-9}\times 10\times 10^{-9}}{(1.8\times 10^{-2})^{2}}= 1.39 \times 10^{-3}N$

Force on -10 nC charge due to -15.0 nC charge

$F_{AC} = \frac{kq_{A}q_{C}}{AC^{2}}= \frac{9\times 10^{9}\times 15\times 10^{-9}\times 10\times 10^{-9}}{(3.5\times 10^{-2})^{2}}= 1.102 \times 10^{-3}N$

Now x Component of net Force

$F_{x}=F_{AC} \cos \theta = 1.102 \times 10^{-3}N\times \cos 30.96^{0} = 0.945\times 10^{-3}N$

y Component of net Force

$F_{y}=F_{BC}-F_{AC} \sin \theta = (1.39-1.102\sin 30.96^{0}) \times 10^{-3}N\times = 0.82\times 10^{-3}N$

So Magnitude of net Force

$F_{net}= \sqrt{F_{x}^{2}+F_{y}^{2}}= \sqrt{(0.945\times 10^{-3}N)+(0.82\times 10^{-3}N)^{2}}= 1.25\times 10^{-3}N$

Direction

$\Phi = \tan^{-1}\left ( \frac{0.82\times 10^{-3}}{0.945\times 10^{-3}} \right )= 40.95^{0}$