# Uniform Electric Field Upward Electric Force Equals Weight. HELP

• m_physics
In summary, when an object with a charge of -3.6uC and a mass of .012kg experiences an upward electric force equal in magnitude to its weight, the direction and magnitude of the electric field can be found by setting the weight equal to the electric field force and solving for E. Using the given values, the electric field is found to be -2.78 x 10^-4 N/C. Additionally, when the electric charge on the object is doubled while its mass remains the same, the direction and magnitude of its acceleration can be found by setting the weight equal to the electric field force and solving for a. The correct answers are a) (-3.3 x 10^4 N/C) in the y direction
m_physics
1. An object with a charge of -3.6uC and a mass of .012kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. (a) Find the direction and magnitude of the electric field. (b) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.
2. E=kQ/r(sqrd)
k=8.99 x 10^9 N*m^2/C^2
magnitude of electron's charge (e): 1.60 x 10 ^ -19 C

3. Attempt
W=mg
W= (.012kg)(10)
W=.12N

Upward force equal to magnitude in weight so...
W= W=kQ/r(sqrd)
.12N + (8.99 x 10^9) [(-3.6uC)(1.6 x 10^-19])/(r^2)
Solve for r
r=-431.52

Plug r into E equation to find electric charge.
E=kQ/r(sqrd)
E=(8.99 x 10^9) [(-3.6uC)(1.6 x 10^-9)]/(-431.52^2)
E= -51.7824/-431.52^2
E= -2.78 x 10 ^ -4

I have no idea about part b. Is it always equal to gravity? Is that how they found the answer.

a) (-3.3 x 10^4 N/C) y direction
b) (9.81 m/s^2) y direction

In the electric field force on a charge is equal to E*q
Hence mg = E*q. Now solve for E.

I would first start by acknowledging that this is a complex problem and it is understandable that you are struggling with it. It is important to break down the problem into smaller, more manageable parts and use the given information and equations to solve for the unknown variables.

In this case, we are given the charge and mass of the object and we know that it experiences an upward electric force equal to its weight. From this, we can use the equation for weight (W=mg) to solve for the magnitude of the electric force, which is also equal to the magnitude of the electric field.

a) To find the direction and magnitude of the electric field, we can use the given information and the equation W=kQ/r^2. Substituting in the values, we get:

W = kQ/r^2
0.12N = (8.99 x 10^9 N*m^2/C^2)(-3.6 x 10^-6 C)/r^2

Solving for r, we get r = -431.52 m. Since r cannot be negative, we can take the absolute value and we get the distance from the charge to be 431.52 m.

Now, we can plug this value back into the equation for electric field (E=kQ/r^2) to find the magnitude of the electric field. This gives us:

E = (8.99 x 10^9 N*m^2/C^2)(-3.6 x 10^-6 C)/(431.52 m)^2
E = 3.3 x 10^4 N/C

Since the electric force and electric field are in the same direction, the direction of the electric field is also in the y direction.

b) To find the direction and magnitude of the acceleration, we can use the equation F=ma and the given information that the mass remains the same. We know that the electric force is equal to the weight, so:

F = ma = 0.12N
Substituting in the mass (0.012 kg), we get:
0.012 kg * a = 0.12N
Solving for a, we get a = 9.81 m/s^2

Since the acceleration is in the same direction as the electric force and electric field, the direction of acceleration is also in the y direction.

In summary, to solve this problem, we used

## 1) What is a uniform electric field and how does it relate to upward electric force and weight?

A uniform electric field is a steady and constant electric field that has the same strength and direction at every point. When an object with a net charge is placed in a uniform electric field, it experiences an upward electric force that is equal in magnitude to its weight. This is because the electric field exerts a force on the object that is opposite to the direction of the gravitational force, resulting in a balance between the two forces.

## 2) How is the upward electric force calculated in a uniform electric field?

The upward electric force can be calculated using the equation F = qE, where F is the force in Newtons, q is the charge of the object in Coulombs, and E is the strength of the electric field in Newtons per Coulomb. This equation follows from the definition of electric force, which is the product of an object's charge and the electric field at its location.

## 3) Can the upward electric force ever be greater than an object's weight?

In a uniform electric field, the upward electric force is always equal to an object's weight. However, in non-uniform electric fields, the electric force can vary in strength and direction at different points, potentially resulting in a force greater than an object's weight. This is known as electrostatic levitation, where an object with a net charge can be suspended in mid-air due to the repulsive force of the electric field.

## 4) How does the direction of the electric field affect the upward electric force?

The direction of the electric field determines the direction of the electric force on an object. In a uniform electric field pointing upwards, the electric force will also be upwards and equal to the weight of the object. If the electric field is pointing downwards, the electric force will be downwards and result in the object being pulled towards the ground.

## 5) What are some real-world examples of a uniform electric field producing an upward electric force equal to weight?

One example is the Van de Graaff generator, which creates a uniform electric field between a metal sphere and a grounded metal plate. Objects with a net charge placed on the sphere will experience an upward electric force equal to their weight, causing them to levitate. Another example is a charged particle in Earth's atmosphere, where the downward electric force due to gravity is balanced by the upward electric force from the Earth's electric field, resulting in the particle remaining suspended in the air.

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