A ${\mathbf{\Pi }}^{\mathbf{-}}$(“pi-minus”) particle, which has charge ${\mathbf{e}}^{-}$, is at location $\mathbf{⟨}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{,}\mathbf{-}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{⟩}\mathbf{\text{\hspace{0.17em}m}}$.

(a) What is the electric field at location $\mathbf{⟨}\mathbf{-}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{,}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{,}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{⟩}\mathbf{\text{\hspace{0.17em}m}}$, due to the π − particle?

(b) At a particular moment an antiproton (same mass as the proton, charge${\mathbf{e}}^{-}$ ) is at the observation location. At this moment what is the force on the antiproton, due to the ${\mathbf{\Pi }}^{\mathbf{-}}$?

The given data can be listed below,

• The location of${\mathrm{\Pi }}^{-}$ particle is, $\mathrm{A}=⟨7\times {10}^{-9},-4\times {10}^{-9},-5\times {10}^{-9}⟩\text{\hspace{0.17em}m}$.
• The location of electric field is, $\mathrm{B}=⟨-5\times {10}^{-9},5\times {10}^{-9},4\times {10}^{-9}⟩\text{\hspace{0.17em}m}$.

The electric force between charged things is described mathematically by Coulomb's law. The electric charge of an item, rather than its mass, determines the amount and sign of the electric force.

The electric field is given by,

$\mathrm{E}=\frac{\mathrm{Kq}}{{\left|\overrightarrow{\mathrm{r}}\right|}^2}$

Here, q is the charge on the particle whose value is $-1.6\times {10}^{-19}\text{\hspace{0.17em}C}$, K is the coulomb constant whose value is$9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2$and $\left|\overrightarrow{\mathrm{r}}\right|$is the magnitude of the distance vector.

The $\overrightarrow{\mathrm{r}}$ is the distance between source and observation location that is given by,

$\overrightarrow{\mathrm{r}}=\mathrm{B}-\mathrm{A}$

Here, B is the observation location and A is the source location.

Substitute values in the above,

$\begin{array}{c}\overrightarrow{\mathrm{r}}=⟨-5\times {10}^{-9},5\times {10}^{-9},4\times {10}^{-9}⟩\text{\hspace{0.17em}m}-⟨7\times {10}^{-9},-4\times {10}^{-9},-5\times {10}^{-9}⟩\text{\hspace{0.17em}m}\\ =⟨-12\times {10}^{-9},9\times {10}^{-9},9\times {10}^{-9}⟩\text{\hspace{0.17em}m}\end{array}$

The magnitude of distance vector is given by,

$\left|\overrightarrow{\mathrm{r}}\right|=\sqrt{{\mathrm{r}}_{\mathrm{x}}^2+{\mathrm{r}}_{\mathrm{y}}^2+{\mathrm{r}}_{\mathrm{z}}^2}$

Substitute all the values in the

$\begin{array}{c}\left|\overrightarrow{\mathrm{r}}\right|=\sqrt{(-12\times {10}^{-9})+(9\times {10}^{-9})+(9\times {10}^{-9})}\text{\hspace{0.17em}m}\\ =17.49\times {10}^{-9}\text{\hspace{0.17em}m}\end{array}$

Substitute all the values in the electric field equation

$\begin{array}{c}\mathrm{E}=\frac{(9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}{{(17.49\times {10}^{-9}\text{\hspace{0.17em}m})}^2}\\ =4.70\times {10}^6\text{\hspace{0.17em}N/C}\end{array}$

Thus, the electric field due to${\mathrm{\Pi }}^{-}$ particle is $4.70\times {10}^6\text{\hspace{0.17em}N/C}$.

The force exerted on antiproton is given by,

$\mathrm{F}=\mathrm{Eq}$

Here, E is the electric field and q. is the charge on antiproton.

$\begin{array}{c}\mathrm{F}=(4.70\times {10}^6\text{\hspace{0.17em}N/C})(1.6\times {10}^{-19}\text{\hspace{0.17em}C})\\ =7.52\times {10}^{-13}\text{\hspace{0.17em}N}\end{array}$

Thus, the force on antiproton is $7.52\times {10}^{-13}\text{\hspace{0.17em}N}$.