Question

What is the force $\overrightarrow{\mathrm{F}}$on the $5.0\mathrm{nC}$charge in FIGURE P$22.43$? Give your answer as a magnitude and an angle measured cw or ccw (specify which) from the $\mathrm{x}$-axis.

Short answer

The value of force is $20.2\times {10}^{-5}\mathrm{N}$and the angle is${45}^{°}$.

Step-by-step solution

Force for y-axis

Force is,

${\mathrm{F}}_{\mathrm{y}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{r}}^2}\mathrm{sin\theta }$

$=9\times {10}^9\times \frac{\left(10\times {10}^{-9}\mathrm{C}\right)\left(5\times {10}^{-9}\mathrm{C}\right)}{{\left(5\times {10}^{-2}\mathrm{m}\right)}^2}\times \frac{4}{5}$

${\mathrm{F}}_{\mathrm{y}}=14.4\times {10}^{-5}\mathrm{N}(-\mathrm{j})$

Calculation for force on x axis

Force is,

${\mathrm{F}}_{\mathrm{x}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\left[\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{{\left({\mathrm{r}}_{32}\right)}^2}-\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\left({\mathrm{r}}_{13}\right)}^2}\mathrm{cos\theta }\right]$

$=9\times {10}^9\times \left[\frac{\left(5\right)\mathrm{nc}\left(5\right)\mathrm{nc}}{9\mathrm{cm}}-\frac{3}{5}\times \frac{\left(10\right)\mathrm{nc}\left(5\right)\mathrm{nc}}{25\mathrm{cm}}\right]\times \frac{{10}^{-18}}{{10}^{-4}}$

${\mathrm{F}}_{\mathrm{x}}=14.2\times {10}^{-5}\mathrm{N}\left(\mathrm{i}\right)$

Figure for force in all axis

Calculation of overall force

Force is,

${\mathrm{F}}_{\mathrm{net}}=\sqrt{{\mathrm{F}}_{\mathrm{x}}^2+{\mathrm{F}}_{\mathrm{y}}^2}$

$=\sqrt{(14.2\mathrm{N}{)}^2+(14.4\mathrm{N}{)}^2}\times {10}^{-5}$

${\mathrm{F}}_{\mathrm{net}}=20.2\times {10}^{-5}\mathrm{N}$

Angle is,

$\mathrm{\theta }={\tan }^{-1}\left(\frac{-14.4\mathrm{N}}{14.2\mathrm{N}}\times \frac{{10}^{-3}}{{10}^{-3}}\right)$

$=-45°$