An object carries a charge of -8.0 ìC, while another carries a charge of -2.0 ìC. How many electrons must be transferred from the first object to the second so that both objects have the same charge?
Qavg = (object 1 + object 2)/2
Qavg= -5ìC
Object 2- Qavg = 3ìC
3E-6 C/1.6E-19 C = 1.875E13 electrons
Did I do this right?
Those "i" things before the "C" were supposed to be micro for microcoulombs
(-2.0 μC+Ne = -8.0 μC-Ne)
2Ne= 6E-6
Ne= 3E-6
N= q/e =3E-6/1.6E-19
N= 1.9E13 electrons
Yes, you have done it correctly. The average charge after transferring electrons would be -5 ìC, and the difference between the final charge of object 2 and the average charge is 3 ìC. To find the number of electrons transferred from object 1 to object 2, you can divide the charge difference by the charge of a single electron (1.6 x 10^-19 C/electron). This gives you 1.875 x 10^13 electrons.
Yes, you have correctly determined that 1.875 x 10^13 electrons must be transferred from the first object to the second object so that both objects have the same charge. Allow me to explain the steps you followed:
1. You calculated the average charge by finding the average of the charges of both objects. In this case, Qavg = (-8 μC + (-2 μC))/2 = -10 μC / 2 = -5 μC.
2. Next, you found the difference between the charge of the second object and the average charge. In this case, Object 2 - Qavg = -2 μC - (-5 μC) = -2 μC + 5 μC = 3 μC.
3. Since the elementary charge of a single electron is 1.6 x 10^-19 coulombs (C), you divided the difference in charge by the elementary charge to find the number of electrons transferred. In this case, (3 μC) / (1.6 x 10^-19 C) = 1.875 x 10^13 electrons.
Therefore, to equalize the charges, 1.875 x 10^13 electrons need to be transferred from the first object to the second.