# Net electric force on neg charge next to a pos charge, inbetween a elec field

• hopeless123
In summary, we have two point charges, Q1 = -6.9 µC and Q2 = 1.1 µC, located between two oppositely charged parallel plates. The distance between the charges is x = 0.38 m and the electric field produced by the plates is uniform and equal to E = 72000 N/C. Using the equations Felectric = K (Q1 * Q2) / (r^2) and Felectric = qE, we find that the net electrostatic force on Q1 is -0.0232 N, directed towards the +vely charged plate.
hopeless123

## Homework Statement

Two point charges, Q1 = -6.9 µC and Q2 = 1.1 µC are located between two oppositely charged parallel plates, as shown in Fig. 16-65. The two charges are separated by a distance of x = 0.38 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 72000 N/C. Calculate the net electrostatic force on Q1 and give its direction.

## Homework Equations

Felectric = K (Q1 * Q2) / (r^2)

Felectric = qE

## The Attempt at a Solution

I'm unsure how to approach from this point. I found the force between the two charges.

F = (9x10^9) [(6.9 x 10^-6)(1.1 x 10^-6) / (.38 ^2)] = .4730609418 N

Then i figured that because it is a constant electric field I could use this equation.
Since the problem states it wants the net charge on Q1, i used that charge for this equation.
F = qE = (6.9 x 10^-6)(72000) = .4968 N

Now I'm trying to find the netforce and whatever I put in seems to be wrong. Electric fields flow from positive to negative so I figured adding these two values would be the correct answer. The force of attraction to Q2 to the right and the force of the electric field to the right. What am I doing wrong?

hopeless123 said:

## Homework Statement

Two point charges, Q1 = -6.9 µC and Q2 = 1.1 µC are located between two oppositely charged parallel plates, as shown in Fig. 16-65. The two charges are separated by a distance of x = 0.38 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 72000 N/C. Calculate the net electrostatic force on Q1 and give its direction.

## Homework Equations

Felectric = K (Q1 * Q2) / (r^2)

Felectric = qE

## The Attempt at a Solution

I'm unsure how to approach from this point. I found the force between the two charges.

F = (9x10^9) [(6.9 x 10^-6)(1.1 x 10^-6) / (.38 ^2)] = .4730609418 N

Then i figured that because it is a constant electric field I could use this equation.
Since the problem states it wants the net charge on Q1, i used that charge for this equation.
F = qE = (6.9 x 10^-6)(72000) = .4968 N

Now I'm trying to find the netforce and whatever I put in seems to be wrong. Electric fields flow from positive to negative so I figured adding these two values would be the correct answer. The force of attraction to Q2 to the right and the force of the electric field to the right. What am I doing wrong?
You have the right equations and the right idea - E fields can be summed linearly - however, Q1 has -ve charge and so will be attracted to the +ve plate, and also to (+vely charged) Q2. So the forces are in opposite directions and one should subtract them to find the net force.

## 1. What is the net electric force on a negative charge placed next to a positive charge in an electric field?

The net electric force on a negative charge placed next to a positive charge in an electric field depends on the strength and direction of the electric field, as well as the distance between the two charges. It can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

## 2. How does the direction of the electric field affect the net electric force on the negative charge?

The direction of the electric field determines the direction of the net electric force on the negative charge. If the electric field is directed towards the positive charge, the force on the negative charge will be in the opposite direction, pulling it towards the positive charge. If the electric field is directed away from the positive charge, the force on the negative charge will also be directed away, pushing it away from the positive charge.

## 3. What happens to the net electric force on the negative charge if the positive charge is moved closer?

If the positive charge is moved closer to the negative charge, the net electric force on the negative charge will increase. This is because the electric field strength is stronger when the charges are closer together, resulting in a greater force between them.

## 4. How does the presence of other charges in the electric field affect the net electric force on the negative charge?

The presence of other charges in the electric field can affect the net electric force on the negative charge. If there are more positive charges in the electric field, the net force on the negative charge may be greater since there are more positively charged particles pulling on it. Conversely, if there are more negative charges in the electric field, the net force on the negative charge may be less since there are more negatively charged particles pushing it away.

## 5. Is there a minimum distance between the positive and negative charge for the net electric force to be zero?

Yes, there is a minimum distance between the positive and negative charge for the net electric force to be zero. This distance is known as the equilibrium distance and can be calculated using Coulomb's law. At this distance, the attractive force between the two charges is equal to the repulsive force, resulting in a net force of zero on the negative charge.

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