Question

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

Answer 1

Answer:

b.  $\Delta KE = 390 eV$

Explanation:

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

$E =\frac{\lambda}{2\pi \epsilon_0 r}$

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

$\Delta V = \int E.dr$

now we have

$\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr$

now we have

$\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})$

now plug in all values in it

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

$\Delta V = 216ln6 = 387 V$

now we know by energy conservation

$\Delta KE = q\Delta V$

$\Delta KE = (e)(387V) = 387 eV$