Answer to Question #91809 in Electric Circuits for Shali

Question #91809
Three charges q1 = + 6.0 µC, q2 = + 9.0 µC and q3 = + 12.0 µC are located at (0, 0) m, (0.3, 0) m and (0.3, 0.4) m respectively. Find the electric force on q3 due to q1, q3 due to q2, find the net force on q3. Show results in unit vector notation.?
1
Expert's answer
2019-07-23T15:53:42-0400


The electric force on q3 due to q1

F13=kq1q3(0.5)2F_{13}=k\frac{q_1 q_3}{(0.5)^2}F13=9109610612106(0.5)2=2.592106(N)F_{13}=9\cdot 10^{9}\frac{6\cdot 10^{-6}\cdot 12\cdot 10^{-6}}{(0.5)^2}=2.592\cdot 10^{-6} (N)

The electric force on q3 due to q1

F13=kq2q3(0.4)2F_{13}=k\frac{q_2 q_3}{(0.4)^2}F13=9109910612106(0.4)2=6.075106(N)F_{13}=9\cdot 10^{9}\frac{9\cdot 10^{-6}\cdot 12\cdot 10^{-6}}{(0.4)^2}=6.075\cdot 10^{-6} (N)

The net force on q3

FN=F132+F232+2F13F23cos(α)F_N=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}cos(\alpha)}

cos(α)=35cos(\alpha)=\frac{3}{5}

FN=(2.592106)2+(6.075106)2+22.5921066.07510635F_N=\sqrt{(2.592\cdot 10^{-6})^2+(6.075\cdot 10^{-6})^2+2\cdot 2.592\cdot 10^{-6}\cdot 6.075\cdot 10^{-6}\cdot\frac{3}{5}}

FN=7.907106(N)F_N=7.907\cdot 10^{-6} (N)


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