Asked by Anon
Three point charges are placed along a straight line with successive charges being 3.0 m apart. The central one has a charge of −3.0 μC, whilst the outer two have charges of +3.0 μC each. If the charges are separated by a vacuum, calculate the magnitude of the total force on one of the outermost charges due to the other two.
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How I am proceeding:
Using 9*10^9 (|3*10^-6|*|3*10^-6|/3^2 + |3*10^-6|*|3*10^-6|/6^2)
But my answer is way out.
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How I am proceeding:
Using 9*10^9 (|3*10^-6|*|3*10^-6|/3^2 + |3*10^-6|*|3*10^-6|/6^2)
But my answer is way out.
Answers
Answered by
bobpursley
why do you have absolute signs on the charges. The sign of the charge is important.
force=kq(-q)/3^2 + kqq/6^2= kqq(-1/9+1/36)=kqq(-4/36+1/36)=kqq*3/36
where q=3e-6
force=kq(-q)/3^2 + kqq/6^2= kqq(-1/9+1/36)=kqq(-4/36+1/36)=kqq*3/36
where q=3e-6
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