Question

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

Answer 1

Answer:

Step-by-step explanation:

U=qv

V= Ed

U=qEd

=(2)(4)(8)=

Answer 2

C

Step-by-step 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