\(\overrightarrow{E_0}=\overrightarrow{E_1}+\overrightarrow{E_2}\) <=> \(\overrightarrow{E_1}=-\overrightarrow{E_2}\Rightarrow\left\{{}\begin{matrix}\overrightarrow{E_1}\uparrow\downarrow\overrightarrow{E_2}\\E_1=E_2\end{matrix}\right.\)

\(E_1=k.\frac{\left|q_1\right|}{r_{10}^2}=k.\frac{\left|q_1\right|}{x^2}\)

\(F_2=k.\frac{\left|q_2\right|}{r_2^2}=k.\frac{\left|q_2\right|}{\left(20-x\right)^2}\)

Có F₀ = F₀ => \(\frac{k\left|q_1\right|}{x^2}=\frac{k.\left|q_2\right|}{\left(20-x\right)^2}\)

<=> \(\frac{16.10^{-8}}{x^2}=\frac{25.10^{-8}}{\left(20-x\right)^2}\)

=> x \(\approx\) 8,89 (cm)

=> E₀ = 2,02.10^-5 (V/m)

Đáp án A.

A = q U M N = -1.2 = - 2 (J). 

Đáp án A.

A = q U M N = -1.2 = - 2 (J). 

Chọn đáp án A.

-2,0 J

Chọn đáp án A.

Chọn đáp án A.

a.Vì q₁ > 0 mà chúng đẩy nhau nên q₂ > 0 

F= \(\frac{k.\left|q_1q_2\right|}{r^2}\)

\(\Rightarrow\left|q_2\right|=\frac{F.r^2}{\left|q_1\right|}=\frac{6,75.10^{-5}.0,02^2}{\left|4.10^{-8}\right|}=0,675\left(C\right)\)

=>q₂ =0,675 C

b) 

b) \(E_{q_1}=\frac{k.\left|q_1\right|}{BH^2}=\frac{9.10^9.\left|4.10^{-8}\right|}{0,01^2}=3,6.10^6\frac{V}{m}\)

\(E_{q_2}=\frac{k.\left|q_2\right|}{AH^2}=\frac{9.10^9.\left|0,675\right|}{0,01^2}=6,075.10^{13}\frac{V}{m}\)

Vì vecto E₁ ↑↑ vecto E₂=>E=|E₁-E₂|=6,075.10¹³ V/m 

\(E_{q_3}=\frac{k.\left|q_3\right|}{AH^2}=\frac{9.10^9.\left|-2.10^{-8}\right|}{\left(0,02.\sin45^o\right)^2}=621,5.10^3\frac{V}{m}\)

Vì vecto E vuông góc với Eq3 nên:

E_H =\(\sqrt{E_{q_3}^2+E^2}=6,075.10^{13}\left(\frac{V}{m}\right)\)