Repulsion between two pith balls

[Biya]

i kinda need help. okay I am posting my question:
Two pith balls have a mass of 1.0g and have equal charges. One pith ball is suspended by an insulating thread. The other is attached to a static insulator and is brought close to the suspended ball so that they are 5 cm apart when repulsion occurs. After repulsion, they come to rest such that both are at the same height above ground. The suspended ball is now hanging with the thread forming a 30 degree angle with respect to its former position. Calculate the following:
1-Force of weight of the suspended ball.
2-The force of repulsion between the two.
3-The net charge on each.
I have no idea how to find out the force of repulsion and the net charge. We can find the weight my using the W=mg formula. So I'm confused with that. I need help . Thank God i found this cool forum

 

[Integral]

Start by drawing a force diagram. Since both balls are stationary you forces must all balance.

 

[wais]

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Hi there!

First of all, don't worry, we are here to help you with your question. Let's break it down step by step.

1. Force of weight of the suspended ball:

As you correctly mentioned, we can use the formula F=mg to calculate the force of weight. In this case, the mass of the ball is 1.0g, and the acceleration due to gravity (g) is approximately 9.8 m/s^2. Therefore, the force of weight of the suspended ball is:

F = (1.0g)(9.8 m/s^2) = 0.0098 N

2. Force of repulsion between the two pith balls:

To calculate the force of repulsion, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The equation is:

F = k(q1q2)/d^2

where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q1 and q2 are the charges of the two pith balls, and d is the distance between them.

In this case, we know that the distance between the two balls is 5 cm, which is equal to 0.05 m. We also know that the charges of the two balls are equal. Therefore, the equation becomes:

F = k(q^2)/d^2

Now, we need to find the value of q (the charge on each ball). To do this, we can use the fact that the balls are at the same height after repulsion, which means that the force of repulsion is equal to the force of weight of the suspended ball. Therefore, we can equate the two equations and solve for q:

F = F = k(q^2)/d^2

0.0098 N = (9x10^9 Nm^2/C^2)(q^2)/(0.05 m)^2

Solving for q, we get:

q = 1.4x10^-8 C

Now, we can plug this value of q into the equation for force of repulsion to find the answer:

F = (9x10^9 Nm^2/C^2)((1.4x10^-8 C)^2)/(0.