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(B) Initially,the bead is at rest,which implies that the upward electric force $F_e = qE$ balances the downward gravitational force $mg$. Thus,$qE = m

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the bead moves downwards and then moves upwards

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(B) Initially,the bead is at rest,which implies that the upward electric force $F_e = qE$ balances the downward gravitational force $mg$. Thus,$qE = mg$.When the electric field is switched off,the only force acting on the bead is gravity $(mg)$. The bead will start moving downwards with a constant acceleration $g$.When the electric field is switched on again,the electric force $F_e = qE$ acts upwards again. Since the bead has gained some downward velocity during the time the field was off,it will continue to move downwards while decelerating until its velocity becomes zero.After the velocity becomes zero,the electric force $F_e$ (which is equal to $mg$) will cause the bead to accelerate upwards until it returns to its original position.Therefore,the bead moves downwards and then moves upwards.

$A$ charged bead is sliding freely through a string held vertically under tension. An electric field is applied parallel to the string so that the bead stays at rest at the middle of the string. If the electric field is switched off momentarily and switched on again,then

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