In Fig. 21-38, particle 1 of charge $\mathbf{+}\mathbf{4}\mathbf{e}$is above a floor by distance ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{2.00}\mathbf{\text{mm}}$and particle 2 of charge$\mathbf{\text{+6e}}$is on the floor, at distance${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{6.00}\mathbf{\text{mm}}$horizontally from particle 1.What is the xcomponent of the electrostatic force on particle 2 due to particle 1?

1. Particle 1 of charge$+4e$ is above a floor by distance,${\mathrm{d}}_1=2\text{\hspace{0.17em}mmor}0.002\text{\hspace{0.17em}m}$
2. Particle 2 of charge$+6e$ is on the floor, at distance${\mathrm{d}}_2=6\text{\hspace{0.17em}mmor}0.006\text{\hspace{0.17em}m}$ horizontally from particle 1.

Using the given charges in the formula of force from Coulomb's law, we can get the x-component of force on particle 2 due to particle 1. To get the x-component of the force, we can get the unit-vector of the distance.

Formulae:

The magnitude of the electrostatic force between any two particles,

${\mathrm{F}}_{21}=\frac{\left|{\mathrm{q}}_1{\mathrm{q}}_2\right|}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2},\mathrm{where}\mathrm{r}=\sqrt{{\mathrm{d}}_1^2+{\mathrm{d}}_2^2}$ (i)

The distance vector in unit-vector notation, $\widehat{\mathrm{r}}=\frac{\mathrm{r}}{\overrightarrow{\mathrm{r}}}=\frac{{\mathrm{d}}_2\widehat{\mathrm{i}}-{\mathrm{d}}_1\widehat{\mathrm{j}}}{\sqrt{{\mathrm{d}}_1^2+{\mathrm{d}}_2^2}}$ (ii)

Since both$q_1$and$q_2$are positively charged, particle 2 is repelled by particle 1, so the direction of ${\mathrm{F}}_{21}$is away from particle 1 and toward 2.

In unit-vector notation, the force acting on the particle 2 due to 1 can be given as:$\overrightarrow{{\mathrm{F}}_{12}}={\mathrm{F}}_{12}\widehat{\mathrm{r}}$

Now, for the x-component of the force, we can take the $\hat{i}$vector value of the distance from equation (ii) as given:

$\begin{array}{c}{\overrightarrow{\mathrm{F}}}_{12\left(\mathrm{x}\right)}=\frac{{\mathrm{F}}_{12}{\mathrm{d}}_2}{\sqrt{{\mathrm{d}}_1^2+{\mathrm{d}}_2^2}}\\ =\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2{\mathrm{d}}_2}{{({\mathrm{d}}_1^2+{\mathrm{d}}_2^2)}^{3/2}}\text{(substitutingequation(i))}\\ =\frac{(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}})(4\times 1.60\times {10}^{-19}\text{\hspace{0.17em}C})(6\times 1.60\times {10}^{-19}\text{\hspace{0.17em}C})(6.00\times {10}^{-3}\text{\hspace{0.17em}m})}{{[{(2.00\times {10}^{-3}\text{\hspace{0.17em}m})}^2+{(6.00\times {10}^{-3}\text{\hspace{0.17em}m})}^2]}^{\frac{3}{2}}}\\ =1.31\times {10}^{-22}\text{\hspace{0.17em}N}\end{array}$

Hence, the x-component of the force is$1.31\times {10}^{-22}\text{\hspace{0.17em}N}$