Question

What is the force$\overrightarrow{\mathrm{F}}$on the$1.0\mathrm{nC}$charge at the bottom in FIGURE $\mathrm{P}22.45$$?$Give your answer in component form.

Short answer

Force on$1.0\mathrm{nC}$is${\overrightarrow{\mathrm{F}}}_{\text{on}1}=1.1\times {10}^{-5}\hat{\mathrm{j}}\mathrm{N}$.

Step-by-step solution

Figure for force on the1.0 nccharge

Figure is,

Calculation of force on q2andq3

Putting a charge of$1.0\mathrm{nC}$at the origin and labelling it${\mathrm{q}}_1$.

For charge${\mathrm{q}}_2$:

${\overrightarrow{\mathrm{F}}}_2\text{on}1=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\hat{\mathrm{i}}$

$=\frac{\left(9\times {10}^9\right)\left(1\times {10}^{-9}\right)\left(2\times {10}^{-9}\right)}{{\left(5\times {10}^{-2}\right)}^2}\left[-\cos 45°\hat{\mathrm{i}}-\sin \left(45°\right)\hat{\mathrm{j}}\right]$

$=\left(0.72\times {10}^5\mathrm{N}\right)\left[-\cos 45°\hat{\mathrm{i}}-\sin 45°\hat{\mathrm{j}}\right]$

For chargelocalid="1649089752680" ${\mathrm{q}}_3$:

${\overrightarrow{\mathrm{F}}}_{3\text{on}1}=\frac{\left(9\times {10}^9\right)\left(1\times {10}^{-9}\right)\left(6\times {10}^{-9}\right)}{{\left(5\times {10}^{-2}\right)}^2}\hat{\mathrm{j}}\mathrm{N}$

$=2.16\times {10}^{-5}\hat{\mathrm{j}}\mathrm{N}$

Force for chargeq4andq1

For charge ${\mathrm{q}}_4$:

${\overrightarrow{\mathrm{F}}}_{4\text{on}1}=0.72\times {10}^{-5}\mathrm{N}\left[\cos 45°\hat{\mathrm{i}}-\sin 45°\hat{\mathrm{j}}\right]$

For charge ${\mathrm{q}}_1$:

${\overrightarrow{\mathrm{F}}}_{\text{on}1}={\overrightarrow{\mathrm{F}}}_{2\text{on}1}+{\overrightarrow{\mathrm{F}}}_{3\text{on}1}+{\overrightarrow{\mathrm{F}}}_{4\text{on}1}$

${\overrightarrow{\mathrm{F}}}_{\text{on}1}=1.1\times {10}^{-5}\hat{\mathrm{j}}\mathrm{N}$