Solve Two Physics Problems: Ebonite Rod & Electroscope

AI Thread Summary
To solve the first physics problem, the charge on the pith ball can be calculated by dividing the total excess electrons (6.4 x 10^8) by the charge of a single electron (-1.602 x 10^-19 coulombs), resulting in a charge of approximately 1.02 x 10^-10 coulombs. For the second problem, the Earth acquires a charge equal to the total charge of the electrons that leave the electroscope, which can be calculated by multiplying the number of electrons (2.5 x 10^11) by the charge of an electron, yielding a charge of approximately -4.00 x 10^-2 coulombs. Both problems illustrate fundamental concepts of charge transfer and conservation in electrostatics. Understanding these calculations is essential for mastering basic physics principles.
Skipperchrldr
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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?
 
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The charge of an electron is -1.602 x 10^-19 coloumbs.

cookiemonster
 

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