A charge q1 = 2.92 uC is at a distance d= 1.13 m from a second charge q2 = -5.77

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A charge q1 = 2.92 uC is at a distance d= 1.13 m from a second charge q2 = -5.77uC

Calendar-glackboardh xa chegg psychologyìnA/Xe-Cengage yi Phys tg0B-Homework 3 secure i https://www webassign.net/web/Student Assignment Responses/submit?dep 1943982 Subenit Answer Save Progress Practice Another Version 9. -1 points 16 19.3.WA 029.Tutorial A charge q, 2.92 uC is at a distance d 1.13 m from a second charge q2- -5.77 c. My Notes Ask 91 t a point A between the two charges that is 2d/3 from q1. Note that the location in the diagram above is not to scale. (b) Find a point between the two charges on the horizontal ntal line where the electric potential is zero. (Enter your answer as measured from q) where Additional Materials Reading Tutorial 10. 1 points OSColPhys2016 19.4 WA 033. Tutorial My Noles Ask Your 231 PM O Type here to search

Explanation / Answer

Electric potential is a scalar quantity, which is given by:

V = kQ/R

Part A.

Vnet = V1 + V2

Vnet = kq1/r1 + kq2/r2

r1 = 2d/3 = 2*1.13/3 = 0.753 m

r2 = d/3 = 1.13/3 = 0.377 m

q1 = 2.92 uC & q2 = -5.77 uC

Using these values

Vnet = 9*10^9*2.92*10^-6/0.753 + 9*10^9*(-5.77*10^-6)/0.377

Vnet = -102844.96 V = -1.03*10^5 V

Part B.

Suppose at x = r from q1, betwwen both charges net electric potential is zero, then

Vnet = V1 + V2 = 0

V1 = -V2

kQ1/R1 = -kQ2/R2

Q1/R1 = -Q2/R2

R1 = r, then R2 = d - r = 1.13 - r

So

2.92/r = -(-5.77)/(1.13 - r)

2.92*(1.13 - r) = 5.77*r

2.92*1.13 = 5.77*r + 2.92*r

r = 2.92*1.13/(5.77 + 2.92)

r = 0.3797 m = 0.38 m

Electric potential is zero at x = 0.38 m from charge q1

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