kaiwangster777
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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

Respuesta :

skyluke89
1) The electric field between the two plates is [tex]E=800.0 V/m[/tex]. The potential difference between the plates is given by
[tex]\Delta V= Ed[/tex]
where d is the separation between the plates. By using 
[tex]d=2.5 cm=0.025 m[/tex]
we find
[tex]\Delta V = Ed = (800.0 V/m)(0.025 m)=20 V[/tex]

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:
[tex]W=\Delta U = e \Delta V[/tex]
where e is the electron charge. By substituting numbers, we find
[tex]W=e \Delta V = (1.6 \cdot 10^{-19}C)(20 V)=3.2 \cdot 10^{-18} J[/tex]

Answer:

1). 20v

2) 3.2

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