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Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00 cm away from the line of charge and released from rest. What kinetic energy in eV will it have when it reaches a position of 5.0 cm away from the initial position? (Note: use the potential difference equation for an infinite line.) a. 350 eV b. 390 ev c. 350k ev d. 6.2 eV e. 170 e

Respuesta :

Answer:

b.  [tex]\Delta KE = 390 eV[/tex]

Explanation:

As we know that the electric field due to infinite line charge is given as

[tex]E =\frac{\lambda}{2\pi \epsilon_0 r}[/tex]

here we can find potential difference between two points using the relation

[tex]\Delta V = \int E.dr[/tex]

now we have

[tex]\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr[/tex]

now we have

[tex]\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})[/tex]

now plug in all values in it

[tex]\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})[/tex]

[tex]\Delta V = 216ln6 = 387 V[/tex]

now we know by energy conservation

[tex]\Delta KE = q\Delta V[/tex]

[tex]\Delta KE = (e)(387V) = 387 eV[/tex]

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