Answer to Question #254341 in Electricity and Magnetism for Vikbah

Question #254341
Three charges q1=q2=+6 and q3=4 are located at - 0.6i, 0.6i and 0.8j respectively. Find the net force on q3
1
Expert's answer
2021-10-21T16:35:57-0400

The net force on the third charge is obtained from the forces between each couple of charges:


F1=kq1q3/r2=kq1q3/((0.6)2+0.82),F1=2.21011.F2=kq2q3/r2=kq1q3/(0.62+0.82),F2=2.21011.F_1=kq_1q_3/r^2=kq_1q_3/((-0.6)^2+0.8^2),\\ F_1=2.2·10^{11}.\\ F_2=kq_2q_3/r^2=kq_1q_3/(0.6^2+0.8^2),\\ F_2=2.2·10^{11}.\\

The angle between the forces is


θ=180°2tan10.60.8=106°.\theta=180°-2\tan^{-1}\frac{0.6}{0.8}=106°.

The magnitude of the resultant is


FR=F12+F222F1F2cosθ=3.11010.F_R=\sqrt{F_1^2+F_2^2-2F_1F_2\cos\theta}=3.1·10^{10}.

It is oriented in +j direction.


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