91影视

As two cosmic rays, one with a charge $q_1$and the other with a charge $q_2$, are separated by a radius $r$, they exert forces upon one another, with the force diminishing as the distance grows and increasing as the magnitude of the charges increases. This force is described by Coulomb, which is expressed as equation ($22.2$) in the form

Equation $1$

$F=\frac{K\left|q_1\right|\left|q_2\right|}{r^2}$

Where $K$is the electrostatic constant, which is equal to $9.0脳{10}^9\mathrm{N}脳{\mathrm{m}}^2/{\mathrm{C}}^2脳q_1$The charge of the metal sphere is $q_1$, while the charge of the ball is $q_2$.To derive the form, solve equation ($1$) for $q_2$.

$F=\frac{K\left|q_1\right|\left|q_2\right|}{r^2}$

Equation $2$

$q_2=\frac{r^{2}F}{K{q}_1}$

The distance between the sphere's centre and the ball is called $r$. As a result, it might be calculated using

$r=\frac{25\mathrm{cm}}{2}+50\mathrm{cm}=62.5\mathrm{cm}$

The number of electrons on the metal sphere is $1脳{10}^{13}$, and each photon's charge is$-1.6脳{10}^{19C}$, and thus the total charges on the ball are

$q_1=\left(1脳{10}^{13}\right)\left(鈭�1.6脳{10}^{鈭�19}\mathrm{C}\right)=鈭�1.6脳{10}^{鈭�6}\mathrm{C}$

To find the magnitude of the charge $q_2$on the ball, put the values for $q_1$, $F$, $r$, and $k$ into equation ($2$).

$q_2=\frac{r^{2}F}{K{q}_1}$

$=\frac{{\left(62.5脳{10}^{鈭�2}\mathrm{m}\right)}^2\left(9.2脳{10}^{鈭�4}\mathrm{N}\right)}{\left(9.0脳{10}^9\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(鈭�1.6脳{10}^{鈭�6}\mathrm{C}\right)}$

$=鈭�25脳{10}^{鈭�9}\mathrm{C}$

The charge of $q_2$is

$$\begin{gathered} q_2=\left(鈭�25脳{10}^{鈭�9}\mathrm{C}\right)\left(\frac{\mathrm{nC}}{1脳{10}^{鈭�9}\mathrm{C}}\right) \\ =鈭�25\mathrm{nC} \end{gathered}$$