The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
Question Id: PH0000006 Question : The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Answer : (a) Distance between the two spheres is 0.12m. (b) Force on the second sphere due to the first sphere is save and i.e 0.2 N. Explanation : Given, two small charge \(q_{1}=0.4\mu C=0.4\times 10^{-6}C=4\times 10^{-7}C\) , \(q_{2}=-0.8\mu C=-0.8\times 10^{-6}C=-8\times 10^{-7}C\) and force(F) between them is 0.2 N. Distance between the sphere of charges is r (say) Figure-1: force between 0.4 µC and –0.8 µC According to Coulomb's law If two point charge \(q\) and \(Q\) are separated by a distance \(r\) then electrostatic force acting on them is \(F=\frac{1}{4\pi \epsilon }\frac{qQ}{r^{2}}\) ;[where \(\epsilon\) is electric permittivity ...