Question

Two parallel plates are set up with one 4.60 m above the other. The bottom plate has an electric potential of zero and the top has an electric potential of 3.39×10⁶ V. A charged ball of mass 1.40 g and charge 6.50 nC is shot from the bottom plate to the top plate at a velocity of 3.30 m/s. What will be the final height of the ball between the plates?
0.412 m
0.573 m
1.80 m
3.21 m

Answer 1

answer) the question gives us a following

V=Ed

E=V/d=3.39*10⁶/4.6=7.34*10⁵V/m

here E will be negative since its travelling downward

and acceleration can be found out by

a=-g+qE/m=-9.8m/s²+6.5*10^-9C*-7.34*10⁵V/m/0.0014kg=-13.21m/s²

now we have

a*(h-0)=1/2*(0-v)²

-13.21*h=-5.445

h=0.412m

so the answer is 0.412m