Aproton is released from rest at the origin in a uniform electric field that is directed in the positive x direction with magnitude 950 \text{ n/c}950 n/c. what is the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 \text{ m}x=2.5 m

Explanation:

It is given that,

Electric field,

We need to find the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 m

From the conservation of energy, the loss in potential energy is equal to the gain in kinetic energy and kinetic energy is is equal to the work done.

So, the change in electric potential energy of the proton field system is . Hence, this is the required solution.

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