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
Answer:
1
The arrow with greater impart is Arrow B
2
The both arrows will feel the same impulse
Explanation:
1. Arrow B since 
it used more force to stop itself in a shorter distance.
2. They should feel the same impulse since the both had the same momentum 
Answer:
Atomic name is your answer.
Answer: the answer should be 6,720 decameters.
Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2
=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons
