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The liquid drop of mass m has a charge q. What should be the magnitude of electric field E to balance this drop ?(A) mg/q (B) E/m (C) mgq (D) mq/g

Answer»

udent,◆ ANSWER -(A) E = mg/q● Explaination -Consider a LIQUID drop of MASS m and charge q.For ELECTRIC field to balance liquid drop, gravitational force and electrostatic force should be equal.Fg = Femg = qEE = mg/qTherefore, magnitude of electric field is mg/q.Thanks for asking. HOPE this is helpful...



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