Question

The magnitude of the electric field between two parallel charged plates is 800.0 . An electron moves to the negative plate 2.5 cm away.

Find the electric potential difference and the work. Recall that the charge of an electron is
1.602 × 10–19 C.

ΔV =
____V

W =
______ × 10–18 J

Answer 1

1). 20v

2) 3.2

Answer 2

1) The electric field between the two plates is
`E=800.0 V/m`. The potential difference between the plates is given by

`\Delta V= Ed`
where d is the separation between the plates. By using

`d=2.5 cm=0.025 m`
we find

`\Delta V = Ed = (800.0 V/m)(0.025 m)=20 V`

2) The work done by the electric field to move the electron by that distance is equal to the variation of electric potential energy of the electron:

`W=\Delta U = e \Delta V`
where e is the electron charge. By substituting numbers, we find

`W=e \Delta V = (1.6 \cdot 10^(-19)C)(20 V)=3.2 \cdot 10^(-18) J`