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3 months ago

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

itude 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

1 answer:

Ostrovityanka [3.2K]3 months ago

7 0

The alteration in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

In the query, it is stated that

The intensity of the uniform electric field equals $E = 950 \ N/C$

The distance the electron covers is $x = 2.50 \ m$

Typically, the force exerted on this electron is expressed mathematically as

$F = qE$

Where F signifies the force and q represents the charge of the electron, which is a fixed 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 is mathematically framed as

$W = \Delta KE$

Where W denotes the work done on the electron by the electric field and $\Delta KE$ is the change in kinetic energy

Additionally, work done on the electron can also be described as

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

Where $\theta = 0 ^o$ assuming that the electron's movement aligns with the x-axis

So

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

Inserting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

According to the conservation of energy

$\Delta PE = - \Delta KE$

Where $\Delta PE$ signifies the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

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