Answer to Question #274124 in Electricity and Magnetism for Abdul

Question #274124

Two fixed charges 1.07 micro Coulomb and -3.29micro Coulomb are 61.8cm apart where may a third charge be located so that no net force act on it?

need better explantion please


1
Expert's answer
2021-12-02T10:06:09-0500

The third charge should be placed at the point, where two electric fields due to charges compensate each other (equal). Let d=61.8cm=0.618md=61.8cm = 0.618m be the distance between charges, q1=1.07×106C,q2=3.29×106Cq_1 = 1.07\times 10^{-6}C, q_2 = -3.29\times 10^{-6}C charges, xx - distance from the first charge. Then, at the point of zero resultant force have:


kq1x2=kq2(d+x)2k\dfrac{|q_1|}{x^2}=k\dfrac{|q_2|}{(d+x)^2}

The equation for xx is then:


q1(d2+2dx+x2)=q2x2(q1q2)x2+2dq1x+q1d2=0|q_1|(d^2+2dx+x^2) = |q_2|x^2\\ (|q_1|-|q_2|)x^2+2d|q_1|x + |q_1|d^2=0

Substituting numbers and solving the quadratic equation for xx, obtain:


x0.82mx\approx 0.82m

Answer. The third charge can be located at 0.82m from the first one.


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