A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength $$81$$$$π$$$$7$$×$$105$$ $$V/m$$. When the field is switched off, the drop is observed to fall with terminal velocity $$2$$×$$10$$−$$3$$ $$m/s$$. Given $$g=9.8$$ $$m/s2$$, viscosity of the air $$1.8$$×$$10$$−$$5$$ $$Ns/m2$$ and the density of oil $$900kgm-3$$ . The magnitude of q is:

Question

A tiny spherical oil drop carrying a net charge q  is balanced in still air with a vertical uniform electric field of strength $$81$$$$π$$$$7$$×$$105$$ $$V/m$$. When the field is switched off, the drop is observed to fall with terminal velocity $$2$$×$$10$$−$$3$$ $$m/s$$. Given $$g=9.8$$ $$m/s2$$, viscosity of the air $$1.8$$×$$10$$−$$5$$ $$Ns/m2$$  and the density of oil $$900kgm-3$$ . The magnitude of  q is:

Infinity Learn · Accepted Answer

GivenElectric field strength $$E=81$$$$π$$$$7$$×$$105$$ VTerminal velocity  $$Vt=2$$×$$10$$−$$3$$ $$m/sAcceleration$$ due to gravity  $$g=9.8$$ $$m/s2Viscosity$$ of the air $$=1.8$$×$$10$$−$$5$$ $$Ns/m2$$ The density of oil  ρ$$oil=900$$ $$kg/m3When$$ the electric field is on then oil drop is in $$equilibriumqE=mg$$∵$$m=43$$$$π$$$$R3$$ρ⇒$$qE=43$$$$π$$$$R3$$ρg⇒$$q=4$$$$π$$$$R3$$ρ$$g3E$$…..(1)When$$ the electric field is switched off $$mg=6$$πηRvt⇒$$43$$$$π$$$$R3$$ρ$$g=6$$πηRvt⇒$$R=9$$η$$vt2$$ρg…..$$(2)From$$ equation $$(1) and (2) we have $$q=43$$$$π$$$$9$$η$$vt2$$ρ$$g32$$×ρgE⇒$$q=43$$$$π$$$$9$$×$$1.8$$×$$10$$−$$5$$×$$2$$×$$10$$−$$32$$×$$900$$×$$9.83/2$$×$$900$$×$$9.8$$×$$781$$π×$$105$$⇒$$q=8$$×$$10$$−$$19C$$

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