A proton moves in a constant electric field E from A to pointB. The magnitude of the elctric field is 4.2x10^4 n/c; Thedirection opposie to that of the proton. If the distance from pointA to point B is 0.18m, what is the change in the proton's electricpotential energy, EPEA-EPEB?
Force (retarding) is F = Q E = 1.6*10^-19 * 4.2*10^4 Newtons
It is losing kinetic energy and gaining potential energy
Work = F d = F * .18 Joules
To calculate the change in the proton's electric potential energy between points A and B, we can use the formula:
ΔEPE = q * ΔV
Where:
ΔEPE is the change in electric potential energy
q is the charge of the proton
ΔV is the change in electric potential
Since we know the magnitude of the electric field E and the distance d between points A and B, we can first calculate the change in electric potential ΔV using the formula:
ΔV = -Ed
Where:
E is the magnitude of the electric field
d is the distance between the points A and B
Substituting the given values:
E = 4.2 × 10^4 N/C
d = 0.18 m
Now we can calculate ΔV:
ΔV = - (4.2 × 10^4 N/C) * (0.18 m)
Next, to find the change in electric potential energy ΔEPE, we need to know the charge of the proton. The charge of a proton is +1.6 × 10^-19 coulombs.
Substituting the values:
q = +1.6 × 10^-19 C
Finally, we can calculate ΔEPE:
ΔEPE = (1.6 × 10^-19 C) * [ΔV calculated earlier]
By performing the calculation, you can find the change in the proton's electric potential energy between points A and B.