A 2 g object,located in a region of uniform | vedclass

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(C) Given: Mass $m = 2 \ g = 2 	imes 10^{-3} \ kg$.Electric field $E = 300 \ NC^{-1}$.Initial kinetic energy $(KE)_1 = 0$ at $x = 0$.Final kinetic en

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$800$

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(C) Given: Mass $m = 2 \ g = 2 	imes 10^{-3} \ kg$.Electric field $E = 300 \ NC^{-1}$.Initial kinetic energy $(KE)_1 = 0$ at $x = 0$.Final kinetic energy $(KE)_2 = 0.12 \ J$ at $x = 0.5 \ m$.According to the work-energy theorem,the work done by the electric force is equal to the change in kinetic energy.Work done $W = F \cdot d = (QE) \cdot d$.Change in kinetic energy $\Delta KE = KE_2 - KE_1 = 0.12 \ J - 0 \ J = 0.12 \ J$.Equating the two: $Q 	imes 300 	imes 0.5 = 0.12$.$Q 	imes 150 = 0.12$.$Q = \frac{0.12}{150} = 0.0008 \ C$.$Q = 800 	imes 10^{-6} \ C = 800 \ \mu C$.Since the object gains kinetic energy moving in the direction of the electric field,the charge $Q$ must be positive.

$A$ $2 \ g$ object,located in a region of uniform electric field $E = (300 \ NC^{-1}) \hat{i}$,carries a charge $Q$. The object is released from rest at $x = 0$ and has a kinetic energy of $0.12 \ J$ at $x = 0.5 \ m$. Then,$Q$ is: (in $\mu C$)

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