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Chọn đáp án C

+ ta có:$\left\{\begin{array}{l}{e}_1=E_0\cos \omega t\\ \left\{\begin{array}{l}{e}_2=E_0\cos \left(\omega t+\frac{2\pi }{3}\right)\\ e_3=E_0\cos \left(\omega t-\frac{2\pi }{3}\right)\end{array}\right.\end{array}\right.\Rightarrow \left|e_2-e_3\right|=E_0\left|\cos \left(\omega t+\frac{2\pi }{3}\right)-\cos \left(\omega t-\frac{2\pi }{3}\right)\right|$

+ Lại có: $\cos \left(\omega t+\frac{2\pi }{3}\right)-\cos \left(\omega t-\frac{2\pi }{3}\right)-2\sin \omega t.\sin \frac{\pi }{3}=-\sqrt{3}\sin \omega t\Rightarrow \left|e_2-e_3\right|=E_0\sqrt{3}\left|\sin \omega t\right|$

+ Theo đề: $\left\{\begin{array}{l}{e}_1=30\Rightarrow E_0\cos \omega t=30\\ \left|e_2-e_3\right|=30\Rightarrow E_0\sqrt{3}\left|\sin \omega t\right|=30\end{array}\right.\Rightarrow {\left(\frac{30}{E_0}\right)}^2+{\left(\frac{30}{E_0\sqrt{3}}\right)}^2={\cos }^2\omega t+{\sin }^2\omega t$.

$\Rightarrow {\left(\frac{30}{E_0}\right)}^2+{\left(\frac{30}{E_0\sqrt{3}}\right)}^2=1\Rightarrow E_0=20\sqrt{3}\left(V\right)\approx 34,64V$