Answer:
V₂ = -22 V
Explanation:
Electric potential and field are related
ΔV = - E d
where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates
In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m
V₂- V₁ = - E d
V₂ = - Ed + V₁
V₂ = - 4000 0.004 + (- 6)
V₂ = -16 - 6
V₂ = -22 V
1). Calculate how long it takes an object to fall 4,000 m after it's dropped. (Use D = (1/2) (g) (T²) . D is 4,000 m. g = 9.8 m/s². Find T .)
2). Calculate how far the object will move HORIZONTALLY in that length of time, if it's moving at 75 m/s. (Distance = (75 m/s) x (time) . )
Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes.
In our case:
Q=2e=2*(-1.6*10^-19) C
V=75 V
Ep=(-3.2*10^-19)*75
Ep=-2.4*10^-17 J
The change in potential energy of the charge is -2.4*10^-17 J