Chrome File Edit View History Bookmarks People Window Help Cengage vy PHY 1520/1

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Chrome File Edit View History Bookmarks People Window Help Cengage vy PHY 1520/1620 Chapter 20: x, : D EGR2400-F17,CassSchedule eta, Calculate The Speed Of A × x x C Owww.webassign.net/web/Student/Assignment-Responses/submit?dep. 1698483 1 #QI . 0/10 points | Previous Answers SerPOPS 20 P.007 (a) Find the potential at a distance of 0.980 cm from a proton. (Note: Assume a reference level of potential V- 0 atr-.) y Notes As Your response differs from the correct answer by more than 10%. Double check your calculations, v (b) What is the potential difference between two points that are 0.980 cm and 2.06 cm from a proton? (c) Repeat parts (a) and (b) for an electron. Need Help? Read in Submit Answer Save ProgressPr acice Another Version Practice Another Version

Explanation / Answer

First the givens:
q = 1.602x10^-19
ke = 8.98755x10^9
r = (0.980/100) <--conversion from cm to m

Electric Potential a distance from a point charge is found by the following formula:
Vq = [1/(4*pi*e0)] q/r

ke is a simplified term for [1/(4*pi*e0)], therefore:

Vq = (ke * q) /r
= (8.98755x10^9* 1.602x10^-19)/9.8e-3=1.469e-7


Next, follow the same method as before, but instead of .980/100 for r, plug in the value: (2.06/100). After you get that answer, take the original (.98 cm) potential and subtract it from (2.06cm) to get the potential difference magnitude.
(8.98755x10^9* 1.602x10^-19)/2.06e-2=6.989e-8
6.989e-8-1.469e-7=7.701e-8


c) Same thing only the field's direction is reversed

i) Vq = (ke * q) /r
= (8.98755x10^9* (-1.602x10^-19))/9.8e-3= -1.469e-7

ii) 6.989e-8-1.469e-7= -7.701e-8

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