Answer to Question #233034 in Physics for meditron

Question

A small sphere 1 carrying charge q₁ and having inertia m₁ is constrained to moving inside a narrow vertical tube Fixed at the bottom of the tube is a small sphere 2 carrying .

.charge q₂

Determine the equilibrium height h

for sphere 1 (ignore friction).

Express your answer in terms of Coulomb's constant k

, acceleration due to gravity g

, and the variables q₁, q₂, and m₁

Answer 1

There are two forces acting on the sphere 1: Coulomb's force acting up and gravitation acting down. The first one at height $h$ is given as follows:

$$F_C = k\dfrac{q_1q_2}{h^2}$$

The second one at any height (small compared with the size of the Earth) is given as follows:

$$F_g = m_1g$$

At equilibrium position they are equal to each other. Thus, obtain for height:

$$F_C = F_g\\$$ $$k\dfrac{q_1q_2}{h^2} = m_1g\\ kq_1q_2 = h^2m_1g\\ h = \sqrt{\dfrac{kq_1q_2}{m_1g}}$$

Answer. $h = \sqrt{\dfrac{kq_1q_2}{m_1g}}$.