# How to find a charge on a sphere in equilibrium?

• sphys4
In summary, the conversation is about finding the charge on a small plastic sphere attached to a string in an electric field. The solution involves calculating the horizontal force on the sphere and using it to find the charge, but the initial attempt was incorrect. The conversation also includes a reminder to include the tension force from the string in the free-body diagram.
sphys4

## Homework Statement

A small, plastic sphere of mass m = 126 g is attached to a string as shown in the figure.

There is an electric field of 151 N/C directed along the + x axis. If the string makes an angle 30 with the y-axis when the sphere is in equilibrium, what is the charge on the sphere?

## The Attempt at a Solution

mg = 0.126kg * 9.81m/s^2 = 1.24 N

To make an angle of 30 degrees, the horizontal force must be 1.24 N * sin30 = 0.62 N

So then i did .62 N/151N/C = .0041 C or 4.1 mC which is the units they want, i entered it but this isn't correct.

Can someone tell me where i went wrong?

sphys4 said:

## The Attempt at a Solution

mg = 0.126kg * 9.81m/s^2 = 1.24 N

To make an angle of 30 degrees, the horizontal force must be 1.24 N * sin30 = 0.62 N
That's not correct.
So then i did .62 N/151N/C = .0041 C or 4.1 mC which is the units they want, i entered it but this isn't correct.

Can someone tell me where i went wrong?
Start by drawing the free-body diagram for the sphere.

How do i incorporate the string in?!

The tension in the string exerts a force on the sphere, so it's another force in the free-body diagram.

## 1. How do I determine the magnitude of the charge on a sphere in equilibrium?

In order to determine the magnitude of the charge on a sphere in equilibrium, you will need to use the equation Q = 4πε₀r²V, where Q is the charge, ε₀ is the permittivity of free space, r is the radius of the sphere, and V is the potential at the surface of the sphere. You can measure or calculate the potential at the surface of the sphere using a voltmeter or by using the electrostatic force equation F = kQq/r² and solving for Q, where k is the Coulomb's constant and q is a known test charge.

## 2. How do I determine the direction of the electric field at a point on the surface of a charged sphere in equilibrium?

The direction of the electric field at a point on the surface of a charged sphere in equilibrium is always perpendicular to the surface. This means that the electric field lines will point directly away from or toward the center of the sphere depending on the sign of the charge. You can also use the right-hand rule to determine the direction of the electric field by pointing your thumb in the direction of the force on a positive test charge and your fingers in the direction of the electric field.

## 3. Can the charge on a sphere in equilibrium ever be negative?

Yes, the charge on a sphere in equilibrium can be negative. This means that the electric field will point inward toward the center of the sphere, rather than outward. The magnitude of the negative charge will be equal to the magnitude of the positive charge, but with the opposite sign.

## 4. How does the distance between two charged spheres affect the charge on each sphere in equilibrium?

The distance between two charged spheres does not directly affect the charge on each sphere in equilibrium. However, as the distance between the spheres decreases, the electric field between them increases, leading to stronger repulsive or attractive forces between the spheres. This can cause the spheres to move and redistribute the charge, ultimately resulting in a new equilibrium state.

## 5. Can a charged sphere ever be in equilibrium without any external influence?

Yes, a charged sphere can be in equilibrium without any external influence if the net electric force on the sphere is zero. This can occur when the sphere is surrounded by other charged objects that create a balanced distribution of electric fields around the sphere. Alternatively, the sphere can be in equilibrium if it is surrounded by a conducting material, which allows the excess charge to distribute evenly across its surface.

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