Answer to Question #267118 in Physics for Godson

Question

Point charges 88μC,-55μC and 70 μC are placed in a straight line. The central one is 0.75m from each of the others. Calculate the net force on each due to the other two.

Answer 1

$q_1=88\ \mu C$ , $q_2=-55\ \mu C$ , $q_3=70\ \mu C$ , $r=0.75\ (m)$

$F_1=k\frac{q_1q_2}{r^2}-k\frac{q_1q_3}{(2r)^2}=\frac{kq_1}{r^2}(q_2-\frac{q_3}{4})=$

$=\frac{9\cdot10^9\cdot 88\cdot10^{-6}}{0.75^2}(55\cdot10^{-6}-\frac{70\cdot10^{-6}}{4})=52.8\ (N)$

$F_2=-k\frac{q_2q_1}{r^2}+k\frac{q_2q_3}{r^2}=\frac{kq_2}{r^2}(-q_1+q_3)=$

$=\frac{9\cdot10^9\cdot 55\cdot10^{-6}}{0.75^2}(-88\cdot10^{-6}+70\cdot10^{-6})=-15.84\ (N)$

$F_3=-k\frac{q_3q_2}{r^2}+k\frac{q_3q_1}{(2r)^2}=\frac{kq_3}{r^2}(-q_2+\frac{q_1}{4})=$

$=\frac{9\cdot10^9\cdot 70\cdot10^{-6}}{0.75^2}(-55\cdot10^{-6}+\frac{88\cdot10^{-6}}{4})=-36.96\ (N)$