# Finding charges on an equilibrium position

• -EquinoX-
These are the weight of the ball (mg) and the electric force (Fe). The ball is at equilibrium, so the net force in this direction must be zero. This means that mg must be equal to Fe. Next, we can look at the forces along the string. These are the tension (T) and the horizontal component of the electric force (Fe cos 37). Since the ball is not moving, the net force in this direction must also be zero. This leads us to the equation T = Fe cos 37. Now we can use the equation E = Fe/q0 to solve for the charge of the ball. Plugging in the given values, we get q0 = (3.00i + 5
-EquinoX-

## Homework Statement

http://img214.imageshack.us/img214/4430/fieldoc3.th.jpg

E = (3.00i+5.00j) x 10^5 N/C the ball is at equilibrium at 37 degrees, the ball has a mass of 1 gram. What is the charge of the ball? What is the tension of the string?

E = Fe/q0

## The Attempt at a Solution

As far as I know to solve for this problem Fe must be equal to 0?

Or we should first resolve the forces acting on the ball, which is

T cos(37) - mg = 0.. am I right? I don't know what's next

Last edited by a moderator:
So in this type of problems, you always treat the horizontal and vertical components separately. Or, in this case, you would use components along and perpendicular to the string.

Let's start by identifying all the forces working perpendicular to the string.

I would approach this problem by first identifying the forces acting on the ball at equilibrium. As mentioned, there are two forces acting on the ball: the tension force from the string and the force of gravity. Using the given information, we can set up the following equations:

T cos(37) - mg = 0
T sin(37) = Eq0

Where T is the tension force, m is the mass of the ball, g is the acceleration due to gravity, E is the electric field, and q0 is the charge of the ball.

Since the ball is at equilibrium, the tension force and the force of gravity must be equal and opposite. We can solve the first equation for T and substitute it into the second equation:

T = mg/cos(37)
T sin(37) = Eq0
mg/cos(37) sin(37) = Eq0
mg tan(37) = Eq0
q0 = mg tan(37)/E

Now, we can solve for the charge of the ball using the given values of m, g, and E. The mass of the ball is given as 1 gram, which is equivalent to 0.001 kg. The acceleration due to gravity on Earth is approximately 9.8 m/s^2. The electric field is given as (3.00i+5.00j) x 10^5 N/C, which can be represented as 3.00 x 10^5 N/C in the x-direction and 5.00 x 10^5 N/C in the y-direction. Therefore, we can calculate the charge of the ball as:

q0 = (0.001 kg)(9.8 m/s^2) tan(37)/(3.00 x 10^5 N/C)
q0 = 2.6 x 10^-8 C

This means that the charge of the ball is 2.6 x 10^-8 Coulombs.

To calculate the tension force, we can substitute the calculated value of q0 into the first equation:

T cos(37) - mg = 0
T cos(37) = mg
T = mg/cos(37)
T = (0.001 kg)(9.8 m/s^2)/cos(37)
T = 1.3 x 10^-5 N

Therefore, the tension force in the string is

## 1. What is an equilibrium position?

An equilibrium position is a point where the net force acting on an object is zero. In other words, the object is either at rest or moving at a constant velocity.

## 2. How do you determine charges on an equilibrium position?

The charges on an equilibrium position can be determined by analyzing the forces acting on the object. If the object is at rest, the net force must be equal to zero, so the magnitudes of the forces must be equal and opposite. These forces can be calculated using 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.

## 3. What factors affect the charges on an equilibrium position?

The charges on an equilibrium position are affected by the distance between the charges, the magnitude of the charges, and the medium through which the charges are interacting. The medium can affect the electric field and thus the forces acting on the charges.

## 4. Can the charges on an equilibrium position change?

Yes, the charges on an equilibrium position can change if the distance between the charges, the magnitude of the charges, or the medium changes. Any changes in these factors can result in a different net force and therefore a different equilibrium position.

## 5. How is the concept of equilibrium position used in real-world applications?

The concept of equilibrium position is used in many real-world applications, such as in designing and testing electrical circuits, understanding the behavior of atoms and molecules, and studying the properties of materials. It is also important in fields like engineering, physics, and chemistry, where the behavior of systems at rest or in motion is of interest.

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