Solving Two Charge Problems - Coulombs and Electrons

[Skipperchrldr]

Hey everyone! I was wondering if someone could help me with these two physics problems i have to turn in tomorrow. Here's the first:
An ebonite rod with an excess of 6.4 X 10^8 electrons shares its charge equally with a pith ball when they touch. What is the charge on the pith ball, in coulumbs?

Here's the second:
How much charge does the Earth acquire if 2.5 x 10^11 electrons leave a grounded metal-leaf electroscope?

 

[Doc Al]

This might help: The charge on an electron is -1.6 x 10^-19 Coulombs.

 

[M98Ranger]

Sure, I can help you with these problems. Let's start with the first one. To find the charge on the pith ball, we can use the equation Q = Ne, where Q is the charge, N is the number of electrons, and e is the elementary charge. We are given that the ebonite rod has an excess of 6.4 x 10^8 electrons, so N = 6.4 x 10^8. Plugging this into the equation, we get Q = (6.4 x 10^8)(1.6 x 10^-19) = 1.024 x 10^-11 coulombs. Therefore, the charge on the pith ball is 1.024 x 10^-11 coulombs.

Moving on to the second problem, we can use the same equation to find the charge acquired by the Earth. We are given that 2.5 x 10^11 electrons leave the electroscope, so N = 2.5 x 10^11. Plugging this into the equation, we get Q = (2.5 x 10^11)(1.6 x 10^-19) = 4 x 10^-8 coulombs. Therefore, the Earth acquires a charge of 4 x 10^-8 coulombs.

Remember, coulombs (C) is the unit of electric charge, and electrons (e) are the fundamental particles that carry negative charge. I hope this helps you with your problems. Good luck with your assignment!