Answer to Question #91809 in Electric Circuits for Shali

Question

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.?

Answer 1

The electric force on q3 due to q1

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

The electric force on q3 due to q1

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

The net force on q3

$$F_N=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}cos(\alpha)}$$

$$cos(\alpha)=\frac{3}{5}$$

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

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