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(A) Given: Mass of the particle $m = 2 	imes 10^{-6} \ kg$,charge on the particle $q = 5 	imes 10^{-6} \ C$.The electric field $E$ due to a charged

Vedclass · Accepted Answer

$35.4 	imes 10^{-12} \ C \ m^{-2}$

Vedclass · Answer

(A) Given: Mass of the particle $m = 2 	imes 10^{-6} \ kg$,charge on the particle $q = 5 	imes 10^{-6} \ C$.The electric field $E$ due to a charged conducting surface is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density.For the particle to hang in the air,the electric force must balance the gravitational force acting on it.$F_e = F_g \Rightarrow qE = mg$Substituting $E = \frac{\sigma}{\epsilon_0}$ into the equation:$q \left( \frac{\sigma}{\epsilon_0} ight) = mg$$\sigma = \frac{mg \epsilon_0}{q}$Substituting the given values:$\sigma = \frac{(2 	imes 10^{-6} \ kg) 	imes (10 \ m \ s^{-2}) 	imes (8.85 	imes 10^{-12} \ C^2 N^{-1} m^{-2})}{5 	imes 10^{-6} \ C}$$\sigma = \frac{20 	imes 8.85 	imes 10^{-18}}{5 	imes 10^{-6}}$$\sigma = 4 	imes 8.85 	imes 10^{-12} \ C \ m^{-2}$$\sigma = 35.4 	imes 10^{-12} \ C \ m^{-2}$

$A$ particle of mass $2 \times 10^{-6} \ kg$ with a charge $5 \times 10^{-6} \ C$ is hanging in air above a similarly charged conducting surface. The charge density of the surface is (Assume $\epsilon_0 = 8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ and $g = 10 \ m \ s^{-2}$)

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