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(A) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.$F_e = mg$Since $F_e = qE$ and $E =

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$2 	imes 10^{-16} 	ext{ kg}$

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(A) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.$F_e = mg$Since $F_e = qE$ and $E = \frac{V}{d}$,we have $F_e = \frac{qV}{d}$.Equating the forces: $mg = \frac{qV}{d}$.Rearranging for mass $m$: $m = \frac{qV}{gd}$.Given values: $q = 1.6 	imes 10^{-19} 	ext{ C}$,$V = 980 	ext{ V}$,$d = 8 	ext{ cm} = 0.08 	ext{ m}$,$g = 9.8 	ext{ m/s}^2$.Substituting the values:$m = \frac{1.6 	imes 10^{-19} 	imes 980}{9.8 	imes 0.08}$$m = \frac{1.6 	imes 10^{-19} 	imes 100}{0.08}$$m = \frac{1.6 	imes 10^{-17}}{0.08} = 20 	imes 10^{-17} 	ext{ kg} = 2 	imes 10^{-16} 	ext{ kg}$.

$A$ small particle carrying a negative charge of $1.6 \times 10^{-19} \text{ C}$ is suspended in equilibrium between two horizontal metal plates $8 \text{ cm}$ apart having a potential difference of $980 \text{ V}$ across them. Find the mass of the particle. $[g = 9.8 \text{ m/s}^2]$

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