Question

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on theplates. The center plane at x = 0.0 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m, with an initial kinetic energy K = 300eV, in the positive x-direction as shown.



In the figure, the kinetic energy of the electron as it reaches plate A is closest to:

+2.4 x 10^-16 J

-3.4 x 10^-16 J

-2.4 x 10^-16 J

+3.4 x 10^-16 J

-2.9 x 10^-16 J

Answer 1

The separation between the two large plates is d = 2.4 m
The uniform electric field between the plates is E = 1500 V/m
The electron is projected from x = 0.0 m,with an initial kinetic energy K = 300 eV
We know that
1 eV = 1.6 * 10^-19 J
or K = 300 * 1.6 * 10^-19 J = 4.8 * 10^-17 J
The force acting on the electron is
F = E * q
where q = 1.6 * 10^-19 C
The work done on the electron to move between the plates is
W = F * d
We know that
W = V * q
or V * q = F * d
or V = (F * d/q)
where V is the potential difference between the plates
Another method:
We know from the relation
K = q * V
or V = (K/q)