A proton is released at the origin in a constant electric field of 850 N/C acting in the positive x-direction. Find the change in the electric potential energy associated with the proton after it travels to x = 2.5 m.
The electric potential decreases by an amount q E * (2.5 m)
q is the electron charge with a plus sign (for the proton charge), in Coulombs. E is the electric field strength, in N/C.
The Potential Energy decreases because the field does work on the proton when moving in the +x axis direction
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To find the change in the electric potential energy associated with the proton, we need to calculate the work done by the electric field in moving the proton from the origin to x = 2.5 m. The work done is equal to the change in electric potential energy.
The formula to calculate the work done by the electric field is given by:
Work = Electric Field Strength x Displacement x Cosine of the angle between the electric field direction and the displacement direction.
In this case, the electric field is acting in the positive x-direction, and the displacement is also in the positive x-direction. Thus, the angle between the electric field direction and the displacement direction is 0 degrees.
Plugging in the given values:
Electric Field Strength (E) = 850 N/C
Displacement (d) = 2.5 m
Angle (θ) = 0 degrees
We can calculate the work done using the formula:
Work = E * d * cos(θ)
Since the angle is 0 degrees, the cosine of 0 is 1. Thus, the work done simplifies to:
Work = E * d
Plugging in the values:
Work = 850 N/C * 2.5 m
Work = 2125 J (joules)
Therefore, the change in electric potential energy associated with the proton is 2125 J.