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revision #328 parent #308 GitHub 7e5b4c1 ← older newer →
ru/3.1.12.md +3 −3
@@ -180,9 +180,9 @@
<div class="solutions">
<p class="TxtSolutions"> <br><br>
a) $mg=\frac{kq^2}{R^2}$ - в положении равновесия $\Rightarrow kq^2=mgR^2$<br><br>
−$F'=F-mg=\frac{mgR^2}{R^2+2Rx+x^2}-mg=\frac{mgR^2-mgR^2-2mgRx-mgx^2}{(R+x)^2}=[x\ll R]\approx-\frac{2mgRx}{(R+x)^2}$<br><br>
−Приближения: $(1+x)^\alpha\approx1+\alpha x$<br><br>
−$F'=-\frac{2mgRx}{(R+x)^2}\approx-2mgRx\frac{1}{R^2}(1-2\frac{x}{R})=-\frac{2mgx}{R}-\frac{4mgx^2}{R^2}=[x\ll R]\approx-\frac{2mgx}{R}$<br><br>
+$F'=F-mg=\frac{mgR^2}{R^2+2Rx+x^2}-mg$$=\frac{mgR^2-mgR^2-2mgRx-mgx^2}{(R+x)^2}=[x\ll R]$$\approx-\frac{2mgRx}{(R+x)^2}$<br><br>
+Приближения: $(1+x)^\alpha$$\approx1+\alpha x$<br><br>
+$F'=-\frac{2mgRx}{(R+x)^2}$$\approx-2mgRx\frac{1}{R^2}(1-2\frac{x}{R})$$=-\frac{2mgx}{R}-\frac{4mgx^2}{R^2}$$=[x\ll R]$$\approx-\frac{2mgx}{R}$<br><br>
б) $\left\{\begin{matrix}3mg=\frac{kq^2}{x^2}
\\
\\ mg=\frac{kq^2}{R^2}
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