Question #312979

Three charges each 5.0x10-6C are situated at he corners of an equilateral triangle of side 1.0m. Find the force that charge q3 experienced due to the other two charges.

Expert's answer

We know that



Force

F1=kq3q1a2F_1=\frac{kq_3q_1}{a^2}


F1=9×109×5×10−6×5×10−612=0.225NF_1=\frac{9\times10^9\times5\times10^{-6}\times5\times10^{-6}}{1^2}=0.225N

F2=kq2q3a2F_2=\frac{kq_2q_3}{a^2}


F2=9×109×5×10−6×5×10−612=0.225NF_2=\frac{9\times10^9\times5\times10^{-6}\times5\times10^{-6}}{1^2}=0.225N

Fnet=F12+F22+2F1F2cos60°F1=F2=FF_{net}=\sqrt{F_1^2+F_2^2+2F_1F_2cos60°}\\F_1=F_2=F

We know that

Fnet=3FF_{net}=\sqrt{3}F

Fnet=3×0.225=0.3897NF_{net}=\sqrt{3}\times0.225=0.3897N


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