Respuesta :
Given:
The equal charges are placed at a distance of
[tex]\begin{gathered} r=3.606\text{ cm} \\ =3.606\times10^{-2}\text{ m} \end{gathered}[/tex]The mass of each charge is,
[tex]\begin{gathered} m=2.307\text{ g} \\ =2.307\times10^{-3}\text{ kg} \end{gathered}[/tex]The acceleration of each charge is,
[tex]a=216.783\text{ m/s}^2[/tex]To find:
The magnitude of the charge on each sphere
Explanation:
The electric force due to the charges supplies the acceleration of the charges. The electric force is,
[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ F=\frac{9\times10^9\times q^2}{(3.606\times10^{-2})^2} \end{gathered}[/tex]Here, k is the Coulomb constant.
We can write,
[tex]\begin{gathered} F=ma \\ \frac{9\times10^9\times q^2}{(3.606\times10^{-2})^2}=2.307\times10^{-3}\times216.783 \\ q^2=\frac{2.307\times10^{-3}\times216.783\times(3.606\times10^{-2})^2}{9\times10^9} \\ q^2=7.23\times10^{-14} \\ q=\pm2.69\times10^{-7}\text{ C} \\ q=\pm0.269\text{ }\mu C \end{gathered}[/tex]Hence, the magnitude of each sphere is,
[tex]\pm0.269\text{ }\mu C[/tex]