Question

What is the force $\overrightarrow{\mathrm{F}}$on the $-10\mathrm{nC}$charge in FIGURE P$22.42$? 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 $7.3\times {10}^{-3}\mathrm{N}$and the angle is$97.9^{°}$.

Step-by-step solution

Calculation for force on charge

Columns law of force is,

${\mathrm{F}}_{\mathrm{C}}={\mathrm{F}}_{1\text{on}2}={\mathrm{F}}_{2\text{on}1}$

$=\frac{\mathrm{K}\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$

Charge of one and two is,

${\overrightarrow{\mathrm{F}}}_{12}=\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\hat{\mathrm{j}}$

$=\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-10\times {10}^{-9}\mathrm{C}\right)\left(8\times {10}^{-9}\mathrm{C}\right)}{(0.01\mathrm{m}{)}^2}\hat{\mathrm{j}}$

$=-7.2\times {10}^{-3}\hat{\mathrm{j}}\mathrm{N}$

Charge on one and three is,

${\mathrm{F}}_{13}=\frac{{\mathrm{Kq}}_1{\mathrm{q}}_3}{{\mathrm{r}}_{13}^2}$

$=\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-10\times {10}^{-9}\mathrm{C}\right)\left(10\times {10}^{-9}\mathrm{C}\right)}{(0.03\mathrm{m}{)}^2}$

$=-1\times {10}^{-3}\hat{\mathrm{j}}\mathrm{N}$

Calculation of force and angle

Force for all charges is,

${\mathrm{F}}_{\text{net}}=\sqrt{{\overrightarrow{\mathrm{F}}}_{13}^2+{\overrightarrow{\mathrm{F}}}_{12}^2}$

$=\sqrt{{\left(-1\times {10}^{-3}\mathrm{N}\right)}^2+{\left(-7.2\times {10}^{-3}\mathrm{N}\right)}^2}$

$=7.3\times {10}^{-3}\mathrm{N}$

Angle for negative direction is,

$\alpha ={\tan }^{-1}\left(\frac{\left(-7.2\times {10}^{-3}\mathrm{N}\right)\hat{j}}{\left(-1\times {10}^{-3}\mathrm{N}\right)\hat{i}}\right)$

$=82.1°$

Angle for the positive direction is,

${\mathrm{\alpha }}^{\text{'}}=180°-\mathrm{\alpha }$

$=180°-82.1°$

$=97.9°$