Question

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx 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

Answer 1

The change in potential energy is
`\Delta PE = - 3.8*10^(-16) \ J`

Step-by-step explanation:

From the question we are told that

The magnitude of the uniform electric field is
`E = 950 \ N/C`

The distance traveled by the electron is
`x = 2.50 \ m`

Generally the force on this electron is mathematically represented as

`F = qE`

Where F is the force and q is the charge on the electron which is a constant value of
`q = 1.60*10^(-19) \ C`

Thus

`F = 950 * 1.60 **10^(-19)`

`F = 1.52 *10^(-16) \ N`

Generally the work energy theorem can be mathematically represented as

`W = \Delta KE`

Where W is the workdone on the electron by the Electric field and
`\Delta KE` is the change in kinetic energy

Also workdone on the electron can also be represented as

`W = F* x *cos( \theta )`

Where
`\theta = 0 ^o` considering that the movement of the electron is along the x-axis

So

`\Delta KE = F * x cos (0)`

substituting values

`\Delta KE = 1.52 *10^(-16) * 2.50 cos (0)`

`\Delta KE = 3.8*10^(-16) J`

Now From the law of energy conservation

`\Delta PE = - \Delta KE`

Where
`\Delta PE` is the change in potential energy

Thus

`\Delta PE = - 3.8*10^(-16) \ J`