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+### Statement
+
+$6.1.14.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$6.1.14.$ What minimum charge $q$ must be fixed at the lowest point of a spherical cavity of radius $R$ so that a small ball of mass $m$ and charge $Q$ is in stable equilibrium at the highest point of the cavity in a gravitational field?
+
+![ For problem 6.1.14 |448x394, 39%](../../img/6.1.14/statement.png)
+
+### Solution
+
+If we solve this problem using the equilibrium condition
+
+$$
+F_k=mg,
+$$
+
+you will get an incorrect answer, because such an equilibrium would be unstable.
+
+To verify stability, it is necessary to displace the ball by a small angle $\alpha\longrightarrow0$; $(\sin{\alpha}\approx\alpha)$ and consider for which $q$ a restoring force will arise.
+
+It should be noted that for small displacements, the Coulomb force can be considered constant
+
+$$
+F_k=\text{const}=\frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}
+$$
+
+![ For problem 6.1.14 |2239x1673, 80%](../../img/6.1.14/6.1.14.png)
+
+Let us consider the projections of forces onto the $Ox$ axis, parallel to the tangent to the sphere at the point of contact with the ball (inclined at an angle $\alpha$ to the horizontal):
+
+$$
+(mg)_x=-mg \cdot \sin{\alpha}
+$$
+
+$$
+(N)_x=0
+$$
+
+$$
+(F_k)_x=F_k \cdot \sin{\alpha /2}
+$$
+
+In this case, the Coulomb force must be significantly greater than $mg$, i.e., the ball will not fall down but can roll down the wall, which means the equilibrium will be stable if the sum of the projections of all forces onto $Ox$ is greater than zero
+
+$$
+F_k \cdot \sin{\alpha /2}-mg \cdot \sin{\alpha}=0
+$$
+
+$$
+F_k=mg \cdot \frac{\alpha}{\alpha /2}
+$$
+
+$$
+\frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}=2mg
+$$
+
+$$
+q=\frac{32\pi\varepsilon_0mgR^2}{Q}
+$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]