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Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards the right hand and has magnetic field of 4.00 N/C .If an ion of charge +2e is released at rest at the left hand plate .What its K.E when reaches the right hand plate?tell the correct answer with explanation..(a)4ev (B) 32ev (C)64ev(D)16ev

Respuesta :

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

     V=4x8=32V

k.e=eV= 2X32=64eV

Answer:

Answer:

Explanation:

U=qv

V= Ed

U=qEd

=(2)(4)(8)=

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