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.

1
Expert's answer
2022-03-21T12:47:33-0400

We know that



Force

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


F1=9×109×5×106×5×10612=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=9×109×5×106×5×10612=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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