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er taied 6 th brwser or enabled for the Sapling Learning website. Adobe Flash Player is required for Sapling Assignments. Please visit this information link for more Re: /2/2018 11:59 PM 0/100 Gradebook Question 7 of 10 Sapling Learning Mapo are fixed in space with a separation of 909 nm. Calculate the electric potensal at Find the magnitude and direction of the electric field at the same point Potenial Number Magnitude of field: Number N/C Direction of field O Toward the proton O Toward the electron O Another direction about us | careers privacy policyterms ofuse contactus heipExplanation / Answer
(a)909/2 = 454.5n=,4.54x10^-7 m
Electric potential due to electron,
V1=k q1/r
=9x10^9x(-1.6x10^-19) /(4.54x10^-7)
= -3.17x10^-3 V
Electric potential due to proton,
V2=k q2/r
=9x10^9x 1.6x10^-19 /(4.54x10^-7)
=3.17x10^-3 V
Total electric potential,V=V1+V2
=(-3.17x10^-3 V +3.17x10^-3 V ) V
=0 V
(b) As they have opposite polarities, their fields ADD anywhere between them. They both have the same charge = 1.6e-19 Coulombs C. The resulting E field will point away from the proton and towards the electron. At the point 909/2 = 454.5nm, the field therefore has magnitude:
k*1.6e-19/454.5e-19^2 + k*1.6e-19/454.5e-19^2 = 14,096 N/C <---------------------
one of them can not be negative because If one was negative, the resultant would be zero and we know that isn't so. If both charges had the same polarity, their fields would exactly cancel at the midpoint because between them, they are in opposite directions (+ charge fields point radially OUT in all directions, - charge fields point radially IN)
Potential: Zero
Magnitude: 14,096 N/C
Direction: Undetermined
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