What is the force$\overrightarrow{\mathrm{F}}$on the$1.0\mathrm{nC}$charge in FIGURE EX $22.20$? Give your answer as a magnitude and a direction.

Figure is,

To determine the force at the$1\mathrm{nC}$charge, we must first determine the force at the positive and negative$2\mathrm{nC}$charges, which can be called$\mathrm{A}$, $\mathrm{B}$,and$\mathrm{C}$,respectively.

So, to urge the forces from charges and$\mathrm{C}$,

we use the approach indicated within the diagram to seek out the$\mathrm{x}$and$\mathrm{y}$components of them.

The y components eliminate out, as illustrated by the direction of the forces.

And need to place the $\mathrm{x}$components together (which is essentially just find the force of the particle so multiply it by the cosine of the angle ).

Answers are given as a magnitude and a direction by the force$\mathrm{F}$on the $1.0\mathrm{nC}$charge.

$\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

${\mathrm{F}}_{\mathrm{A}\to \mathrm{B}}^{\to }={\mathrm{F}}_{\mathrm{A}\to \mathrm{B}}\mathrm{cos\theta }\hat{\mathrm{i}}+{{\mathrm{F}}^{\to }}_{\mathrm{A}\to \mathrm{B}}\stackrel{}{\sin }\mathrm{\theta }\hat{\mathrm{j}}$

${\overrightarrow{\mathrm{F}}}_{\mathrm{A}\to \mathrm{C}}={{\mathrm{F}}^{\to }}_{\mathrm{A}\to \mathrm{C}}\mathrm{cos\theta }\hat{\mathrm{i}}-{{\mathrm{r}}^{\to }}_{\mathrm{A}\to \mathrm{B}}\mathrm{sin\theta }\hat{\mathrm{j}}$

${\overrightarrow{F}}_A={F^{\to }}_{A\to B}\stackrel{}{\cos \theta }\hat{i}+{F^{\to }}_{A\to C}\stackrel{}{\cos \theta \hat{i}}$

${\overrightarrow{\mathrm{F}}}_{\mathrm{A}}=\mathrm{k}\frac{\left(1\times {10}^{-9}\left(2\times {10}^{-9}\right)\right.}{(0.01{)}^2}\cos 60°+\mathrm{k}\frac{\left(1\times {10}^{-9}\left(2\times {10}^{-9}\right)\right.}{(0.01{)}^2}\cos 60°$

${\overrightarrow{\mathrm{F}}}_{\mathrm{A}}=1.789\times {10}^{-4}\mathrm{N}$