Question

A small metal sphere has a mass of 0.15 g and a charge of -21.0 nC. It is 10.0 cm 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

Electric force between 2 charged particles (derived from Coulomb's Law)

F = Q1Q2 / (4πɛr^2)

Q1 = charge on first particle
Q2 = charge on second particle

ɛ = permittivity of free space (a constant, actually epsilon nought, can't do subscript here), 8.85 x 10^-12 F/m

r is the distance between the centres of the 2 particles.

So need to convert everything to SI units.

F = (-21n)(-21n) / [(4πɛ)(0.1^2)]
= 3.965 x 10^-4 N (3 sig fig)

Since charges are negative on both spheres, they experienced a repelling electric force.

Upon release, the sphere is now only under 2 forces, the electric force and its weight. Since it is directly above the lower sphere, the electric force is acting directly upwards, hence negating some of the weight.

Net force on sphere = Weight - Fe (taking downward to be positive)

Weight = mg
= 0.15 x 10^-3 x 9.81
= 1.471 x 10^-3N

Net force on sphere = (1.471 x 10^-3) - (3.965 x 10^-4)
= 1.074 x 10^-3 (Hence net downward force, therefore begins to fall)

Initial acceleration under this force

Fnet = ma
a = 1.074 x 10^-3 / (0.15 x 10^-3)
= 7.16m/s^2