Answer to Question #106255 in Electricity and Magnetism for 888

Question

[img]https://upload.cc/i1/2020/03/23/XIalyt.jpg[/img]



the question is in the picture

Answer 1

Three forces exert on the ball. The force of gravity is directed downward, the Coulomb repulsion force is directed horizontally away from the second ball. The reaction force of the support is directed strictly perpendicular to the surface of the bowl, i.e. along the radius to the center of the bowl. These forces are drawn in the figure. Since the ball is in equilibrium, the sum of all forces acting on it must be equal to 0. The projection of the sum of forces on the directions of the X and Y axes must also be equal to 0.  For Y direction we have

(1) $N_y-P=0$ ,

For X direction we have

(2) $F_k-N_x=0$

Signs in these conditions take into account the directions of force. The magnitudes of forces are 

$P=mg$ ; $F_k=k\frac{q^2}{d^2}$ where k is Coloumb's constant; $N_x=Ncos\alpha; N_y=Nsin\alpha$

Substitute this to the equilibrium conditions (1) and (2) we get two equations.

$Nsin\alpha=mg\\ Ncos\alpha=k\frac{q^2}{d^2}$

Let's exclude an unknown force $N$ from these equations . To do this, divide  second by first

$\frac {cos\alpha}{sin\alpha}=k\frac{q^2}{mgd^2}$

Solving this equation with respect to charges we get

$q=d\sqrt{\frac{mg\cdot cos\alpha}{k\cdot sin\alpha}}$

We find the angle $\alpha$ from considering the geometry of the problem. The triangle AAD is isosceles and its median is height too. Thus we have

$cos\alpha=\frac{d}{2A}; sin\alpha=\sqrt{1-(\frac{d}{2A})^2}$

Answer: The charge of balls is $q=d\sqrt{\frac{mg\cdot cos\alpha}{k\cdot sin\alpha}}$ , where $cos\alpha=\frac{d}{2A}; sin\alpha=\sqrt{1-(\frac{d}{2A})^2}$ , k is  Coulomb's constant, and $g$ - acceleration of free fall.