Question

Charges 5.0×10−7

C,−2.5×10−7

C and 1.0×10−7

C are held fixed at the three corners A,B,C of an equilateral triangle of side 5.0cm respectively. Find the electric force on the charge at C due to the rest two.

Answer 1

$F_1=\frac{kq_Aq_C}{r^2}=\frac{9\cdot 10^9\cdot 5\cdot 1\cdot 10^{-14}}{0.05^2}=0.18~N,$

$F_2=\frac{k|q_B|q_C}{r^2}=\frac{9\cdot 10^9\cdot 2.5\cdot 1\cdot 10^{-14}}{0.05^2}=0.09~N,$

$F=\sqrt{F_1^2+F_2^2+2F_1F_2\cos 120°}=\sqrt{0.09^2+0.18^2-2\cdot 0.09\cdot 0.18\cdot 0.5}=0.16~N.$