Question

Two 2.0 kg masses are 2.0 m apart on a frictionless table. Each has 1.0 μC of charge. What is the magnitude of the electric force on one of the masses? What is the initial acceleration of each mass if they are released and allowed to move?

Answer 1

$From\,\,Coulobm's\,\,law\,\,F=\frac{kq_1q_2}{r^2}\\\\ F\,\,is\,\,force\,\,between\,\,charge\,\,particles\,\,q_1\,\,qnd\,\,q_2\\ Coulomb's\,\,law\,\,constant\,\,k=9\times10^9\,\,N.C^2/m^2\\ Here\,\,q_1=q_2=1.0\,\,\mu C=10^{-6}\,\,C\\ d=2.0\,\,m$

$\therefore \,\,F=\frac{\left (9\times10^9\,\,NC^2/m^2 \right )(1\times10^{-6}\,\,C)^2}{\left (2.0\,\,m \right )^2}\\\\ \Rightarrow \,\,F=2.25\times10^{-3}\,\,N \\\\ \Rightarrow \,\,\left (2.0\,\,kg \right )a=2.25\times10^{-3}\,\,N \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because\,\,F=ma,\,\,\,\,given\,\,m=2.0\,\,kg] \\\\ \Rightarrow \,\,a=\frac{2.25\times10^{-3}\,\,N}{2.0\,\,kg}\\\\ \Rightarrow \,\,a=1.125\times10^{-3}\,\,m/s^2\\\\\\ \therefore \,\,Magnitude\,\,of\,\,the\,\,electric\,\, force\,\,on\,\,one\,\,of\,\,the\,\,masses\,\,F=2.25\times10^{-3}\,\,N \\ And\,\,initial\,\,acceleration\,\,of\,\,each\,\,mass\,\,a=1.125\times10^{3}\,\,m/s^2$

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