A proton is release from rest at the origin in a uniform electric field in the positive x direction with magnitude 850N/C. What is the change in electric potential energy when the proton travels to x = 2.50m
To find the change in electric potential energy, we need to know the formula for electric potential energy and use it to calculate the change when the proton moves from the origin to a point at x = 2.50m.
The formula for electric potential energy is:
ΔPE = q * ΔV
Where:
ΔPE is the change in electric potential energy
q is the charge of the proton
ΔV is the change in electric potential
Since the problem doesn't mention the charge of the proton, we can assume it is the fundamental charge, q = 1.6 x 10^-19 C.
Now, let's find the change in electric potential, ΔV.
The electric potential, V, at a point in the electric field can be calculated using:
V = E * x
Where:
V is the electric potential
E is the electric field strength
x is the distance
In this case, the electric field strength is given as 850 N/C and the distance traveled by the proton is 2.50m. Therefore, we can substitute these values into the equation to get the electric potential at x = 2.50m:
V = 850 N/C * 2.50m
V = 2125 J/C
Now that we have the change in electric potential, we can calculate the change in electric potential energy:
ΔPE = q * ΔV
ΔPE = (1.6 x 10^-19 C) * (2125 J/C)
ΔPE = 3.4 x 10^-16 J
Therefore, the change in electric potential energy when the proton travels to x = 2.50m is 3.4 x 10^-16 J.
ΔKE=W(electric field) =e•Δφ=e•E•x=
=1.6•10⁻¹⁹•850•2.5 =3.4•10⁻¹⁶ J.