  Question

Two identical particles each of mass $$M$$ and charge $$Q$$ are placed a certain distance part. If they are in equilibrium under mutual gravitational and electric force then calculate the order of $$\dfrac{Q}{M}$$ in $$SI$$ units.

Solution

As the given system of charges are in equilibrium,hence $$F_e = F_g$$$$F_e$$ is the electric force between the charges which is repelling them while $$F_g$$ is the gravitational force of attraction.ie. $$\frac{KQ^2}{R^2} = \frac{GM^2}{R^2}$$or , $$KQ^2 = GM^2$$ or , $$(\frac{Q}{M})^2 = (\frac{G}{K})$$or , $$\frac{Q}{M} = ({G}{K})^{\frac{1}{2}}$$as    G = $$6.67 \times 10^{-11} Nm^2 / Kg^2$$and  K = $$9 \times 10^9 Nm^2 / C^2$$$$(\frac{Q}{M})^2 = \frac{( 6.67 \times 10^{-11} )}{ ( 9 \times 10^9 )}$$or $$\frac{Q}{M} = [ 0.7411 \times 10^{-(11+9)} ] ^{\frac{1}{2}}$$ or $$\frac{Q}{M} = ( 0.7411 \times 10^{-20} ) ^{\frac{1}{2}}$$we get $$\frac{Q}{M} = 0.86 \times 10^{-10}$$  c / kgPhysics

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