$3.1.12.$ a. A small charged body of mass $m$ can slide on a vertical spoke, at the lower point of which there is a charge of the same sign as the charge of the body. The equilibrium position of the body is at a distance $R$ from this charge. How does the force acting on the body depend on its small displacement x from the equilibrium position?
b. The mass of the body is tripled, leaving the charges unchanged. At what distance from the lower end of the spoke is the equilibrium position of the body now? How does the force acting on the body depend on its small displacement from the equilibrium position?
a) Equality of forces in the equilibrium position $$ mg=\frac{kq^2}{R^2}\Rightarrow kq^2=mgR^2 $$ $$ F'=F-mg=\frac{mgR^2}{R^2+2Rx+x^2}-mg $$ $$ F'=\frac{mgR^2-mgR^2-2mgRx-mgx^2}{(R+x)^2} $$ Using the approximation for $x\ll R$ $$ F'\approx-\frac{2mgRx}{(R+x)^2} $$ Next we will use approximations: $(1+x)^\alpha $$ \approx1+\alpha x$, где $x\ll1$ $$ F'=-\frac{2mgRx}{(R+x)^2}\approx-2mgRx\frac{1}{R^2}(1-2\frac{x}{R}) $$ $$ F'=-\frac{2mgx}{R}-\frac{4mgx^2}{R^2} $$ Given $x \ll R$, we neglect the term with $R^2$ $$ \boxed{F'\approx-\frac{2mgx}{R}} $$ b) Next, we write down the system of equilibrium equations for the system $$ \left\{\begin{matrix}3mg=\frac{kq^2}{x^2} \\ mg=\frac{kq^2}{R^2} \end{matrix}\right. $$ Now let's find at what distance from the lower end of the spoke the equilibrium position of the body is located. $$ \frac{R}{x}=\sqrt{3}\Rightarrow \boxed{x=\frac{R}{\sqrt{3}}} $$ The force acting on a body depends on its small displacement from the equilibrium position $$ \boxed{F''=-\frac{2m'gx}{R'}=-\frac{6mgx}{R'}}$$
$$F=-\frac{2mgx}{R};\quad R'=\frac{R}{\sqrt{3}};\quad F'=-\frac{6mgx}{R'}$$