# Calculating the magnitude of the net force bewteen charges

• thrills4ever
In summary: To find the x-component of the two balls on the left, you would add the x-component of the right ball and the y-component of the left ball.
thrills4ever

## Homework Statement

Nine different charged balls, which we treat as point charges, are arranged in a highly symmetric pattern around a square. Note that the value of Q is 4.00 x 10-6C, and L = 30.0 cm

Image: http://www.webassign.net/userimages/81224?db=v4net

## Homework Equations

1) What is the magnitude of the net force experienced by the ball with the -3Q charge due to the +3Q and +Q charged balls only?

2) What is the magnitude of the net force experienced by the ball with the -3Q charge due to the four charged balls with charges of -5Q?

3 )What is the magnitude of the net force experienced by the ball with the -3Q charge due to the two charged balls with charges of +2Q?

4) Use your previous three results to find the magnitude of the net force experienced by the ball with the -3Q charge due to the other 8 charged balls.

## The Attempt at a Solution

1) I got this part correct: I understand how to solve for the force when the point charges are in the straight line

My work is as follows:

F3 = F31 + F32
= k(-3Q)(Q) / L^2 + k(3Q)(+3Q) / 4L^2
= 5.25k(Q)^2 / L^2
= 8.39

2) and 3) :

Here is where I do not understand how to solve for the force when there are diagonals involved.

If I am correct in my thinking for 2) then does the -3Q ball experience 2 diagonal forces from the 2 left -5Q balls and a straight force to the right from the 2 right -5Q balls?

In my calculation I did F53 = 15kq^2 / 3L^2, simplifying to 5kq^2 / L^2 which gives me 7.99 I believe. Since there are essentially 2 diagonal forces, do I multiply this by 2 and then add the other force which is directed right by doing 7.99cos(30 degrees)?

Thanks for all the input!

The first part of your reasoning is right.

But consider only the right most column. If we consider this column by itself, why do you think there would be a net force?

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well i tried putting in 0 for the answer and it did not work because I initially thought the opposing diagonals cancel each other out, but apparently that is not the case.

Sorry I meant right, why do you think there would be a force on the right most column?

The two 5q balls on the left do give you a net force that is horizontal -- which can be easily seen if you componentize the forces and use superposition.

Sorry for the mix up.

I see what you mean..im not sure how to use superposition in this case since both the forces from the two 5q balls on the left are in opposing diagonals. So the resultant force is directed to the right horizontally correct? I tried finding the F53x and F53y components but i think that both diagonals produced y components that cancel each other out. I am not sure how to proceed with the question.

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The principal of superposition basically means that the total force is the sum of the forces. You are correct and finding that the y-component of the two balls will cancel out since they are equal and opposite, however you're still left with the x-components.

## What is the formula for calculating the magnitude of the net force between charges?

The formula is F = k * (|q1| * |q2|) / r^2, where F is the net force, k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.

## What is the unit of measurement for the magnitude of the net force between charges?

The unit of measurement is Newtons (N), which is equal to kg*m/s^2.

## How does the distance between charges affect the magnitude of the net force?

The magnitude of the net force is inversely proportional to the square of the distance between charges. This means that as the distance increases, the net force decreases, and vice versa.

## Does the direction of the charges affect the magnitude of the net force?

Yes, the direction of the charges does affect the magnitude of the net force. If the charges are of the same sign (positive or negative), the net force will be repulsive, and if the charges are of opposite signs, the net force will be attractive.

## Can the magnitude of the net force between charges be negative?

No, the magnitude of the net force between charges is always positive. The negative sign is used to indicate the direction of the force, but the magnitude is always positive.

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