Four identical metallic spheres with charges of +1.0 µC, +5.4 µC, −2.6 µC, and −2.0 µC are placed on a piece of paper. The paper is lifted on all corners so that the spheres come into contact with each other simultaneously. The paper is then flattened so that the metallic spheres become separated. a. What is the resulting charge on each sphere?b. How many excess or absent electrons (depending on the sign of your answer to part (a)) correspond to the resulting charge on each sphere?

a) 0.45 μC b) 2.81*10¹² absent electrons.

Explanation:

The total net charge on the 4 spheres, is just the algebraic sum of the individual charges:

Q = Q₁ +Q₂ + Q₃ +Q₄ = +1.0 μC + 5.4 μC -2.6 μC -2.0 μC = +1.8 μC

After being brought in contact each other and then separated, the total charge must be the same as before, i.e., + 1.8 μC.

As the spheres are identical and assumed perfect conductors, at being put in touch each other, there will be a free movement of charges, trying to bring all spheres to a neutral state.

Due to the symmetry of the problem, after being separated, all spheres must carry the same charge, as follows:

Q₁ = Q₂ = Q₃ = Q₄ = Q/4 = +1.8 μC / 4 = +0.45 μC

b) As each sphere has a resulting positive charge, this means that they have a defect of electrons, which number can be found as follows:

c

explanation:

i believe the answer is c and this is only worth 10 points you might wanna change it

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge