Question

A Van de Graaff generator is a device that accumulates electrons on a large metal sphere until the large amount of charge causes sparks. As you’ll learn in Chapter $23$, the electric field of a charged sphere is exactly the same as if the charge were a point charge at the center of the sphere. Suppose that a $25cm$diameter sphere has accumulated $1.0\times {10}^{13}$extra electrons and that a small ball 50 cm from the edge of the sphere feels the force $\overrightarrow{F}=\left(9.2\times {10}^{-4}\mathrm{N}\right.$, away from sphere). What is the charge on the ball?

Short answer

The charge of the ball is$-25nC$.

Step-by-step solution

Equations

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 $9.0\times {10}^9\mathrm{N}\times {\mathrm{m}}^2/{\mathrm{C}}^2\times q_1$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

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

Explanation

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

$q_1=\left(1\times {10}^{13}\right)\left(-1.6\times {10}^{-19}\mathrm{C}\right)=-1.6\times {10}^{-6}\mathrm{C}$

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

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

$=\frac{{\left(62.5\times {10}^{-2}\mathrm{m}\right)}^2\left(9.2\times {10}^{-4}\mathrm{N}\right)}{\left(9.0\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-1.6\times {10}^{-6}\mathrm{C}\right)}$

$=-25\times {10}^{-9}\mathrm{C}$

The charge of $q_2$is

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