Force of two charges

Question

A metal is tested to have a charge of 697.6C. How many electrons are added to it to acquire the charge?

If two similarly charged metals are brought 1.5m apart in a medium of dielectric constant 2, calculate the force between them.

Is the force attractive or repulsive?

Answer 1

I assume the charge is negative, or you would have to remove electrons. In order to find the number of electrons required, you need to divide the charge by the charge per electron (1.60 x 10^-19 C/elec)

Assuming that the metals are small enough to treat them as points compared to the distance apart, 1.5m, so that Coulomb's Law applies, the dielectric constant, κ, affects Coulomb's Law in the following way:

|F| = k_e/κ Q²/d² or 1/(4πκε₀) Q²/d²

Plug in all the knowns and constants to get F in Newtons. The metals are the same charge.

Answer 2

#1 You need to divide the total change by the charge of a single electron.

total # electrons = 697.6 C / 1.602 x 10^-19 C / electron = 4.29 x 10²¹ electrons.

#2 Dielectric constant is defined as the ratio of the permittivity in a medium to the permittivity in a vacuum.

D = ε / ε_0. As D = 2, ε = 2 x ε₀ or ε = 2 x 8.8541 x 10^-12 C²/(N m² = 1.771 x 10^-12.

Using Coulombs law, F = 1/4πε x Q² / 1.5². This assumes the charges are equal in magnitude and of the same sign. The force would be repulsive.