`1.`

Vị trí: `A, B, C` thẳng hàng, `B` nằm giữa `A` và `C` (`AB + BC = AC = 20 \text{ cm}`)

`E_1 = k \frac{|q_1|}{AC^2} = 9 \cdot 10^9 \cdot \frac{12 \cdot 10^{-6}}{0,2^2} = 2,7 \cdot 10^3 \text{ V/m}`

`E_2 = k \frac{|q_2|}{BC^2} = 9 \cdot 10^9 \cdot \frac{3 \cdot 10^{-6}}{0,05^2} = 10,8 \cdot 10^3 \text{ V/m}`

`E_C = E_1 + E_2 = 13,5 \text{ kV/m}`

Đáp án: `13,5`

`2.` 

 Vị trí: `r_1 = r_2 = 6 \text{ cm} = 0,06 \text{ m}`

`E = E_1 + E_2 = \frac{k}{r^2} (|q_1| + |q_2|)`

`E = \frac{9 \cdot 10^9}{0,06^2} (40 \cdot 10^{-6} + 50 \cdot 10^{-6}) = 225 \cdot 10^6 \text{ N/C}`

`225 \cdot 10^6 = 225 \cdot 10^x \Rightarrow x = 6`

Đáp án: `6`

`3.` 

`MB = \sqrt{NB^2 + MN^2} = \sqrt{6^2 + 8^2} = 10 \text{ cm} = 0,1 \text{ m}`

`E_B = k \frac{|q_B|}{MB^2} = 9 \cdot 10^9 \cdot \frac{12 \cdot 10^{-9}}{0,1^2} = 10,8 \cdot 10^3 \text{ V/m} = 10,8 \text{ kV/m}`

`E_A = k \frac{|q_A|}{MA^2} = 9 \cdot 10^9 \cdot \frac{14 \cdot 10^{-9}}{0,08^2} \approx 19,69 \text{ kV/m}`

Vì `\vec{E}_A` $\perp$ `\vec{E}_B` (tại N)

`E_{Mx} = E_B \cdot \cos(\alpha) = 10,8 \cdot \frac{6}{10} = 6,48 \text{ kV/m}`

`E_{My} = E_A + E_B \cdot \sin(\alpha) = 19,69 + 10,8 \cdot \frac{8}{10} = 28,33 \text{ kV/m}`

`E_M = \sqrt{E_{Mx}^2 + E_{My}^2} \approx 29,1 \text{ kV/m}`