Answer to Question #167385 in Physics for Erica

Question

A metal Sphere is given a charge of -3.0X10^-6C and second identical sphere a charge of ＋2.0x10^-6 C.The spheres are separated by distance of 0.010m. (A) what is the force between them? Is this an attractive or repulsive force? (B) The spheres are brought into contact with each other.What happens as a result of this contact? The spheres again separated by 0.010m. What is now the force between them? Is this an attractive or repulsive force?

Answer 1

(a) We can find the force between a metal spheres from the Coulomb's law:

$$F=\dfrac{kq_1q_2}{r^2},$$ $$F=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(-3.0\cdot10^{-6}\ C)\cdot2.0\cdot10^{-6}\ C}{(0.01\ m)^2}=-540\ N.$$

The magnitude of the electrostatic force exerted on one sphere by the other is 540 N. Since the two charges are opposite in sign they will attract each other, therefore, the direction of the electrostatic force is attractive, hence, the force between them is attractive (also, the sign minus indicates that the direction and force are attractive).

b) When the spheres are brought into contact with each other, the charge will shared between the two spheres (each sphere has the same amount of charge):

$$q=\dfrac{q_1+q_2}{2}=\dfrac{-3.0\cdot10^{-6}\ C+2.0\cdot10^{-6}\ C}{2}=-0.5\cdot10^{-6}\ C.$$

c) Let's calculate the force between the two spheres after the separation:

$$F=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(-0.5\cdot10^{-6}\ C)\cdot(-0.5\cdot10^{-6}\ C)}{(0.01\ m)^2}=22.5\ N.$$

The magnitude of the electrostatic force exerted on one sphere by the other is 22.5 N. Since the two charges are both negatively charged they will repel each other, therefore, the direction of the electrostatic force is repulsive, hence, the force between them is repulsive (also, the sign plus indicates that the direction and force are repulsive).