Respuesta :
[tex]\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}[/tex][tex]\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}[/tex][tex]\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}[/tex][tex]\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}[/tex][tex]\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\ \end{gathered}[/tex][tex]\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}[/tex]Hence the correct answer is q= 1.73514 micro coulombs.
