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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 / kg

Physics

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