(#1) If two pith balls - one which is neutral and another which has a charge of -9.6 uC (micro-Coulombs) - are touched together, how many extra electrons will each ball have?
(#2) Charge q1 = 6 uC is at the origin, charge q2 = 7 uC is at y = 2m, x = 0, charge q3 = 1.5 uC is at y = -1 m, x = 0. What is the force on q2?
Pith is an insulator, so there is no reason the two balls would equalize their charges after contact. Only a small fraction would be transfered, and it is hard to say how much. They probably want you to assume the balls acquire equal charges of -4.8 uC after contact.
Divide the charge by the electron charge of 1.6*10^-19 C to get the number of extra electrons.
Your second question has been asked and answered elsewhere.
(#1) To answer this question, we need to understand the concept of charge conservation and the fact that charge is quantized. The charge on an electron is equal to -1.6 x 10^-19 Coulombs (C), which means that each electron has a charge of -1.6 x 10^-19 C.
When two objects are touched together, charge can be transferred between them. In this case, the negatively charged pith ball (-9.6 uC) and the neutral pith ball will come into contact with each other. By touch, charge will flow from the negatively charged ball to the neutral ball until they reach equilibrium.
To find the number of extra electrons gained or lost by each ball, we need to calculate the difference in charges between the two balls.
Given that the charge on the negatively charged ball is -9.6 uC, we can convert it to Coulombs by multiplying by 10^-6:
-9.6 uC x 10^-6 = -9.6 x 10^-12 C
Since the neutral ball has no initial charge, its charge is 0 C.
The difference in charge between the two balls is:
-9.6 x 10^-12 C - 0 C = -9.6 x 10^-12 C
We can now calculate the number of extra electrons gained or lost by each ball. Remember that the charge of each electron is -1.6 x 10^-19 C.
For the neutral ball, the number of extra electrons gained is:
(-9.6 x 10^-12 C) / (-1.6 x 10^-19 C/electron) = 6 x 10^7 electrons gained
For the negatively charged ball, the number of extra electrons lost is:
0 electrons lost since the initial charge is already negative
Therefore, when the two balls are touched together, the neutral ball gains 6 x 10^7 extra electrons, while the negatively charged ball loses 0 extra electrons.
(#2) To find the force on q2, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * (|q1| * |q2|) / r^2
In this case, q1 = 6 uC, q2 = 7 uC, and q3 = 1.5 uC. We need to find the force on q2, which is located at y = 2m, x = 0.
First, we need to find the distance between q2 and q1. Since q2 is at the origin and q1 is at y = 2m, x = 0, the distance between them is:
r1 = 2 m
Next, we need to find the distance between q2 and q3. Since q2 is at y = 2m, x = 0 and q3 is at y = -1 m, x = 0, the distance between them is:
r2 = |2 m - (-1 m)| = 3 m
Now, we can calculate the total force on q2 by considering the forces due to q1 and q3. The force between q2 and q1 is given by:
F1 = k * (|q1| * |q2|) / r1^2
Similarly, the force between q2 and q3 is given by:
F2 = k * (|q2| * |q3|) / r2^2
The total force on q2 is the vector sum of F1 and F2:
F_total = F1 + F2
Using the values provided, we can plug in the numbers and calculate the force on q2.
Note: k is the electrostatic constant and has a value of approximately 9 x 10^9 N m^2 / C^2.