A charged spherical drop of mercury is in equ | vedclass

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(D) For the drop to be in equilibrium,the upward electric force must balance the downward net gravitational force (weight minus buoyancy).$qE = V(\rho

Vedclass · Accepted Answer

$0.95 	imes 10^{-6} \ m$

Vedclass · Answer

(D) For the drop to be in equilibrium,the upward electric force must balance the downward net gravitational force (weight minus buoyancy).$qE = V(ho_{Hg} - ho_{air})g$Since the volume of a sphere is $V = \frac{4}{3}\pi r^3$,we have:$qE = \frac{4}{3}\pi r^3(ho_{Hg} - ho_{air})g$$r^3 = \frac{3qE}{4\pi(ho_{Hg} - ho_{air})g}$Substituting the values: $q = 8 	imes 10^{-18} \ C$,$E = 6 	imes 10^4 \ Vm^{-1}$,$ho_{Hg} = 13.6 	imes 10^3 \ kg/m^3$,$ho_{air} \approx 1.29 \ kg/m^3$,$g = 9.8 \ m/s^2$.$r^3 = \frac{3 	imes 8 	imes 10^{-18} 	imes 6 	imes 10^4}{4 	imes 3.14 	imes (13600 - 1.29) 	imes 9.8}$$r^3 \approx \frac{144 	imes 10^{-14}}{12.56 	imes 13598.71 	imes 9.8} \approx \frac{1.44 	imes 10^{-12}}{1.673 	imes 10^6} \approx 0.86 	imes 10^{-18} \ m^3$$r \approx (0.86 	imes 10^{-18})^{1/3} \approx 0.95 	imes 10^{-6} \ m$.

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