Physics Homework

Question

Two very small spheres are initially neutral and separated by a distance of 0.50m. Suppose that 3.0xe13 electrons are removed from one sphere and put on the other. (A) what is the magnitude of the electrostatic force that acts on each sphere?

Answer 1

GIVEN:

d = 0.50 m.

Q₁ = gains 3 x 10¹³ electrons and the charge of 1 electron = - 1.6 x 10^-19C, thus Q₁ = (3x10¹³)(-1.6x10^-19C) = - 4.8 x 10^-6 C.

Q₂ = loses 3 x 10¹³ electrons thus Q₂ = +4.8 x 10^-6 C.

k = 9 x 10⁹ Nm² / C²

 

UNKNOWN:

F = ?

 

SOLUTION:

Now that we know the charge due to the electron transfer, we can use Coulomb's law to solve for the electrostatic force.

 

F = k (Q₁ Q₂) / d² = 9x10⁹ Nm² /C² (-4.8x10^-6C x +4.8x10^-6C) / (0.50m)² = -0.82944 N or -0.83 N (sig figs), negative because they attract each other.