Question

The Earth's electric field creates a potential that increases 100 V for every meter of altitude. If an object of charge 4.5 mC and mass 68 g falls a distance of 1.0 m from rest under the influence of the Earth's electric and gravitational fields, what is its final kinetic energy

Answer 1

Answer:

The final kinetic energy is  $K = 1.1 \ J$

Explanation:

From the question we are told that

    The electric field is  $E = 100 \ V/m$

    The charge on the object is  $q = 4.5 mC = 4.5 *10^{-3} \ C$

    The mass of the object is  $m_o = 68 \ g = 0.68 \ kg$

     The distance moved by the object is $d = 1.0 \ m$

The workdone on the object by the fields  is  mathematically represented as

   $W = [qE + mg]d$

Now this workdone is equivalent to the final kinetic energy so  

      $K = W = [qE + mg]d$

substituting values

        $K = W = [4.5*10^{-3 } *100 + 0.68 * 9.8]* 1$

        $K = 1.1 \ J$