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

b.
`\Delta KE = 390 eV`

Step-by-step explanation:

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

`E =(\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((\lambda)/(2\pi \epsilon_0 r)).dr`

now we have

`\Delta V = (\lambda)/(2\pi \epsilon_0)ln((r_2)/(r_1))`

now plug in all values in it

`\Delta V = (12* 10^(-9))/(2\pi \epsilon_0)ln((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`