Asked by jenn
A charge of +q is located at the origin, while an identical charge is located on the x axis at x = 0.45 m. A third charge of +4q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?
I am just supposed to be using Coulombs law. I tried in relation to origin charge where the charge at .45 would be F=kq/.45^2 and in relation with charge 4q F-k4q/X^2. Then since q1=q2 the formula was kq/.45^2 = k4q/x^2. simplified that into x^2=4/.45^2. Then took the square root of the answer. it was like 4.444 totally wrong. I cannot figure out this problem at all please some guidance!
I am just supposed to be using Coulombs law. I tried in relation to origin charge where the charge at .45 would be F=kq/.45^2 and in relation with charge 4q F-k4q/X^2. Then since q1=q2 the formula was kq/.45^2 = k4q/x^2. simplified that into x^2=4/.45^2. Then took the square root of the answer. it was like 4.444 totally wrong. I cannot figure out this problem at all please some guidance!
Answers
Answered by
drwls
To double the force at the origin, the +4q charge must have the same effect upon the charge as the +q charge at x=0.45 m
Making the distance x = 0.9 (twice that of the second charge) will result in the same force produced by the +4q particle, because of the the inverse square law.
You were OK with this equation
q/(0.45)^2 = 4q/(0.9)^2
Your mistake was algebra in the next step.
k*q/.45^2 = k*4q/x^2.
leads to
x^2 = 4*(0.45)^2
x = 0.9
Making the distance x = 0.9 (twice that of the second charge) will result in the same force produced by the +4q particle, because of the the inverse square law.
You were OK with this equation
q/(0.45)^2 = 4q/(0.9)^2
Your mistake was algebra in the next step.
k*q/.45^2 = k*4q/x^2.
leads to
x^2 = 4*(0.45)^2
x = 0.9
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