An insulating rod with linear charge density of $$40$$ μ$$C/m$$ and linear mass density of $$0.1$$ $$kg/m$$ is released from rest in a uniform electric field E $$=$$ $$100$$ $$V/m$$ directed perpendicular to the rod. If rod is released and after travelling $$2m$$ down the field, speed of rod is found to be (v/10) $$m/s$$, then value of v is $$______$$

Question

An insulating rod with linear charge density of $$40$$ μ$$C/m$$ and linear mass density of $$0.1$$ $$kg/m$$ is released from rest in a uniform electric field E $$=$$ $$100$$ $$V/m$$ directed perpendicular to the rod. If rod is released and after travelling $$2m$$ down the field, speed of rod is found to be (v/10) $$m/s$$, then value of v is $$______$$

Infinity Learn · Accepted Answer

Taking potential Vinitial $$=$$ $$0$$ at initial position of rod, then potential after travelling distance 'd’ is $$V=-EdCharge$$ on rod $$=$$ λLPotential energy of $$rod-field$$ system initially $$QVinitial=0Potential$$ energy of $$rod-field$$ system $$finally=$$−λLEd $$(using$$, $$U=qV)Now$$ using energy conservation, $$($$$$KE$$$$+$$$$PE$$$$)initial$$ $$=($$$$KE$$$$+$$$$PE$$$$)final$$  We get, $$0+0=12mv02$$−λLEd⇒ $$v02=2$$λ$$LEdm=2$$λLEdμ$$L=2$$λEdμ⇒ $$v0=2$$λEdμ$$=2$$×$$40$$×$$10$$−$$6$$×$$100$$×$$20.1$$ $$=0.400m/s=410m/s$$ So,  $$v=4$$

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