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

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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft