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.?

Expert's answer


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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