Answer to Question #175189 in Physics for dezzzzz

Question #175189


Problem: A metal sphere is given a charge of – 3.0 x 10-6 C and a second identical sphere of 

  +2.0 x 10-6 C. The sphere are separated by a distance of 0.010 meter. (a) What is 

  the force between them? This is attractive or repulsive? (b) the sphere are brought 

  into contact with each other. What happen as a result of this contact? The sphere is

  again separated by 0.010 m. what is now the  force between them? Is this attractive 

  or repulsive force?  



1
Expert's answer
2021-03-24T19:38:45-0400

(a) We can find the force between the 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\\cdot10^{-6}\\ C)\\cdot2\\cdot10^{-6}\\ C}{(0.01\\ m)^2}=-540\\ N."

The magnitude of the electric force between the spheres is 540 N. Since the two charges are opposite in sign they will attract each other, therefore, the direction of the electric force is attractive (also, the sign minus indicates that the direction is 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},""q=\\dfrac{-3\\cdot10^{-6}\\ C+2\\cdot10^{-6}\\ C}{2}=-5\\cdot10^{-7}\\ C."

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


"F=\\dfrac{9\\cdot10^9\\ \\dfrac{Nm^2}{C^2}\\cdot(-5\\cdot10^{-7}\\ C)\\cdot(-5\\cdot10^{-7}\\ C)}{(0.01\\ m)^2}=22.5\\ N."

The magnitude of the electric force between the spheres is 22.5 N. Since the two charges are both negatively charged they will repel each other, therefore, the direction of the electric force is repulsive (also, the sign plus indicates that the direction is repulsive).


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