Question

What is the force$\overrightarrow{\mathrm{F}}$on the$1.0\mathrm{nC}$charge at the bottom in FIGURE $\mathrm{P}22.46$$?$Give your answer in component form.

Short answer

Force on the$1.0\mathrm{nC}$at the bottom is$1.02\times {10}^{-5}\hat{\mathrm{i}}+2.2\times {10}^{-5}\hat{\mathrm{j}}\mathrm{N}$.

Step-by-step solution

Figure for force on the charge1.0 nC

Figure is,

Calculation of forceq2andq3

Force on charge${\mathrm{q}}_2$:

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

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

Force on chargelocalid="1649091309769" ${\mathrm{q}}_3$:

localid="1649091320016" ${\overrightarrow{\mathrm{F}}}_{3\text{on}1}=2.16\times {10}^{-5}\mathrm{N}\hat{\mathrm{j}}$

Calculation for force onq4andq1

Force on charge${\mathrm{q}}_4$:

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

Force on chargelocalid="1649091299979" role="math" ${\mathrm{q}}_1$:

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

$=2\left(0.72\times {10}^{-5}\mathrm{N}\right)\cos \left(45°\right)\hat{\mathrm{i}}+\left(2.16\times {10}^{-5}\mathrm{N}\right)\hat{\mathrm{j}}$

$=1.02\times {10}^{-5}\hat{\mathrm{i}}+2.2\times {10}^{-5}\hat{\mathrm{j}}\mathrm{N}$