Question

A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that has the same charge. This lower sphere is fixed and cannot move. If the upper sphere is released, it will begin to fall.What is the magnitude of its initial acceleration?

Answer 1

a = -7.29 m / s²

Step-by-step explanation:

For this exercise we must use Newton's second law,

F -W = m a

Force is electrical force

F = k q₁ q₂ / r²

k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

q₁ = q₂ = q

a = (k q² / r² - m g) / m

a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward