A -14.8nC point charge and a +23.7nC point charge are 15.0cm apart on the x-axis
ID: 1533719 • Letter: A
Question
A -14.8nC point charge and a +23.7nC point charge are 15.0cm apart on the x-axis. What is the electric potential at the point on the x-axis where the electric field is zero? 6.25x101 v You are correct. Previous Tries Your receipt no. is 164-4795 What is the magnitude of the electric field at the two points on the x-axis where the electric potential is zero? (Input your answers in order of increasing distance from the negative point charge.) 798.08N/C Submit Answer Incorrect. Tries 2/10 Previous Tries 1.50 10 4N/C Submit Answer Incorrect. Tries 1/10 Previous TriesExplanation / Answer
As per the condition -
V1 + V2 = 0
=> kq1/r + kq2/(0.15-r) = 0
=> -14.8/r + 23.7/(0.15-r) = 0
=> -14.8(0.15-r) + 23.7r = 0
=> -2.22 + 14.8r + 23.7r = 0
=> r = 2.22/38.5 = 0.0577 m
Therefore, the requisite electric field, E = kq1/r^2 + kq2/(0.15-r)^2
= -9x10^9 x 14.8 x 10^-9/0.15^2 + 9x10^9 x 23.7 x 10^-9/(0.15-0.0577)^2
= -5920 + 25037
= 19117 N/C
Again, consider the point is not between the charges.
So, in that case -
kq1/r + kq2/(0.15+r) = 0
=> -14.8/r + 23.7/(0.15+r) = 0
=> -14.8(0.15+r) + 23.7r = 0
=> -2.22 - 14.8r + 23.7r = 0
=> r = 2.22/8.9 = 0.25 m
Therefore, the requisite electricfield, E = kq1/r^2 + kq2/(0.15+r)^2
= -9x10^9 x 14.8 x 10^-9/0.15^2 + 9x10^9 x 23.7 x 10^-9/(0.15+0.25)^2
= -5920 + 1333.12
= -4586.87 N/C
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