# Attraction of -Q to +Q in x and y directions

• Violagirl
In summary, the question is asking for the magnitude and direction of the force on the charge -Q in the given figure. The formula for calculating this force is F = k(Q)(-Q)/r^2 * r-hat, where r points from Q to the -Q charge. The attempt at solving the problem includes breaking the force into x and y components, but the directions of the forces need to be checked. The general formula for the force experienced by charge q due to charge Q is F = kQ/r^3 * q * r, where r points from Q to q.

## Homework Statement

Find the magnitude and direction of the force on the charge -Q on the figure in the attached document.

## Homework Equations

F = k(q)(Q)/r2r-hat

## The Attempt at a Solution

So I have a force in the x direction and a force in the y direction.

Fx = k(Q1)(-Q)/(a) x-hat

Fy = (k)(Q2)(-Q)/(a) y-hat

F = Fx + Fy = -kQ22/(a)2 x-hat + -kQ12/a2 y-hat

Is this right? Or can more be done to solve it?

#### Attachments

• Attraction of -Q to +Q in +x and +y directions.jpg
3.4 KB · Views: 404
Almost there - check the directions of the forces.
Your formula has the -Q charge being pushed in the -x direction.
Is that what would happen?

In general - the force experienced by charge q due to charge Q is:
$$\vec{F}=\frac{kQ}{r^3}q\vec{r}$$ Here r points from Q to q.

## 1. What is the concept of attraction of -Q to +Q in x and y directions?

The concept of attraction of -Q to +Q in x and y directions refers to the force of attraction between two electric charges - one negative (-Q) and one positive (+Q) - in both the horizontal (x) and vertical (y) directions. This force is governed by Coulomb's Law, which states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

## 2. How does the distance between the charges affect the attraction in x and y directions?

The distance between the charges has a significant impact on the attraction in x and y directions. As per Coulomb's Law, the force of attraction decreases as the distance between the charges increases. This means that the farther apart the two charges are, the weaker the force of attraction will be in both the x and y directions.

## 3. What is the role of the charges' magnitudes in the attraction in x and y directions?

The magnitudes of the charges play a crucial role in the attraction in x and y directions. As per Coulomb's Law, the force of attraction increases as the magnitude of the charges increases. This means that the larger the magnitude of the charges, the stronger the force of attraction will be in both the x and y directions.

## 4. How does the direction of the charges affect the attraction in x and y directions?

The direction of the charges does not have a significant impact on the attraction in x and y directions. As long as there is one negative (-Q) and one positive (+Q) charge, the force of attraction will exist in both the x and y directions. However, if the charges are both positive or both negative, there will be a repulsive force instead of an attractive force.

## 5. Can the attraction in x and y directions be calculated separately?

No, the attraction in x and y directions cannot be calculated separately. Coulomb's Law applies to the total force of attraction between two charges and does not differentiate between the horizontal and vertical components. However, the total force can be resolved into its x and y components using trigonometry if needed.

• Introductory Physics Homework Help
Replies
25
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
721
• Introductory Physics Homework Help
Replies
16
Views
633
• Introductory Physics Homework Help
Replies
2
Views
126
• Introductory Physics Homework Help
Replies
13
Views
965
• Introductory Physics Homework Help
Replies
7
Views
688
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
889
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
953