Updated spacing between @ latex expressions

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правка #10536 предыдущая #9820 GitHub c702ebe ← раньше позже →
en/7.1.12.md +4 −4
@@ -63,16 +63,16 @@
<h3>Solution</h3>
<p>
Gauss theorem
−$$\oint_{2\pi r}E\,dl=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{q}{2\pi \varepsilon_0 r}$$
+$$\oint_{2\pi r}E\,dl=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{q}{2\pi\varepsilon_0 r}$$
Let's write down the small change of the electrostatic field and then integrate it
$$dU=E \,dr\Rightarrow \int dU=\int E \,dr$$
Integrate from $R_1$ to $R_2$
−$$U_0=\frac{q}{2\pi \varepsilon_0 r}\int_{R_1}^{R_2}\frac{dr}{r}=\frac{q}{2\pi \varepsilon_0 r}\ln\frac{R_2}{R_1}$$
+$$U_0=\frac{q}{2\pi\varepsilon_0 r}\int_{R_1}^{R_2}\frac{dr}{r}=\frac{q}{2\pi\varepsilon_0 r}\ln\frac{R_2}{R_1}$$
Law of conservation of energy
$$\frac{mv^2}{2}=eU$$
−$$e\frac{q}{2\pi \varepsilon_0 \frac{R_1+R_2}{2}}=\frac{mv^2}{(\frac{R_1+R_2}{2})}$$
+$$e\frac{q}{2\pi\varepsilon_0 \frac{R_1+R_2}{2}}=\frac{mv^2}{(\frac{R_1+R_2}{2})}$$
From where we get
−$$\frac{q}{2\pi \varepsilon_0}=\frac{mv^2}{e}$$
+$$\frac{q}{2\pi\varepsilon_0}=\frac{mv^2}{e}$$
Substituting the previously obtained values
$$\frac{U_0}{\ln\frac{R_2}{R_1}}=\frac{2eU}{e}$$
From where
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