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(B) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.$QE = Mg$Since $E = \frac{V}{d}$,we

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$3e$

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(B) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.$QE = Mg$Since $E = \frac{V}{d}$,we have $Q \left( \frac{V}{d} ight) = Mg$.Given $Q = Ne$,where $N$ is the number of elementary charges:$Ne \left( \frac{V}{d} ight) = Mg$$N = \frac{Mgd}{eV}$Substituting the given values ($M = 1.96 	imes 10^{-15} \; kg$,$g = 9.8 \; m/s^2$,$d = 0.02 \; m$,$V = 800 \; V$,$e = 1.6 	imes 10^{-19} \; C$):$N = \frac{1.96 	imes 10^{-15} 	imes 9.8 	imes 0.02}{1.6 	imes 10^{-19} 	imes 800}$$N = \frac{3.8416 	imes 10^{-16}}{1.28 	imes 10^{-16}} = 3$Therefore,the charge on the particle is $Q = 3e$.

Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is:

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