Answer to Question #88589 in Mechanics | Relativity for Tej

Question

In a horizontal uniform electric field, a small charged disk is gently released on the top of a fixed spherical dome. The disk slides down the dome without friction and breaks away from the surface of the dome at angular position ? = sin–1 (3/5) from the vertical. Determine the ratio of the force of gravity acting on the disk to the force of its interaction with the field.

Answer 1

Consider the process of sliding downward with the electromagnetic force F acting vertically upward along the line of force of gravity:

The body starts breaking away from the dome when the net force (which is centripetal force) according to Newton's second law becomes equal to the normal force N. Express the normal force through the force of gravity at a given angle:

$$N=(mg-F)\text{cos}\alpha.$$

The equation of equality of forces:

$$m\frac{v^2_{em}}{R}=(mg-F)\text{cos}\Big(\text{sin}^{-1}\frac{3}{5}\Big),$$

The speed can be calculated with law of conservation of energy:

$$\frac{1}{2}mv_{em}^2=mgR(1-\text{cos}\alpha)=mgR\Big[1-\text{cos}\Big(\text{sin}^{-1}\frac{3}{5}\Big)\Big],$$

$$v_{em}=\sqrt{2gR\Big[1-\text{cos}\Big(\text{sin}^{-1}\frac{3}{5}\Big)\Big]}=\sqrt{0.4gR},$$

substitute this into the Newton's law equation:

$$m\cdot{0.4g}=(mg-F)\cdot0.8,$$

$$\frac{F}{mg}=\frac{1}{2}.$$