# Charge problems and net force

## Homework Statement

A point charge +2Q is at the origin and a point charge -Q is located along the x axis at x=d. Find a symbolic expression for the net force on a third point charge +Q located along the y-axis at y = d

Answer = $$k\frac{Q^2}{d^2}<\frac{1}{2\sqrt{2}}, 2 - \frac{1}{2\sqrt{2}}>$$

Below is a Picasso of what is going on

[PLAIN]http://img708.imageshack.us/img708/842/25027988.jpg [Broken]

## The Attempt at a Solution

I define the charge at the origin, +2Q, to be 1 and the -Q at x = d to be 2 and Q at y = d be 3

$$F_{3} = F_{13} + F_{23}$$

(1) $$F_{13} = k\frac{2q^2}{d^2}<0,1>$$

(2) $$F_{23} = -k\frac{q^2}{2d^2}<-1,1>$$

So the sum of (1) and (2) is then $$k\frac{q^2}{d^2}<\frac{1}{2},-\frac{1}{2}>$$

(1), I had q2 because they use q as a common

(2)I had a - in front of k because it is a negative and positive charge interaction. The <-1,1> indicates that the "slope" of the vector is negative.

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## Answers and Replies

cepheid
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In your second equation (2), your vector components are not right. You need to use a UNIT vector in that direction. Therefore, the magnitude of the vector has to be equal to 1. The vector you used <-1, 1> doesn't have magnitude of 1.

In your second equation (2), your vector components are not right. You need to use a UNIT vector in that direction. Therefore, the magnitude of the vector has to be equal to 1. The vector you used <-1, 1> doesn't have magnitude of 1.

Ohhh

(2) $$F_{23} = -k\frac{q^2}{2\sqrt{2}d^2}<-1,1>$$

cepheid
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Ohhh

(2) $$F_{23} = -k\frac{q^2}{2\sqrt{2}d^2}<-1,1>$$

Close, but not quite. Check your answer. Does the vector $\frac{1}{2\sqrt{2}}<-1,1>$ have a magnitude of 1?

Hint: To get a unit vector in the direction of the vector <-1, 1>, simply take this vector and divide it by its magnitude.

Close, but not quite. Check your answer. Does the vector $\frac{1}{2\sqrt{2}}<-1,1>$ have a magnitude of 1?

Hint: To get a unit vector in the direction of the vector <-1, 1>, simply take this vector and divide it by its magnitude.

In the picture I had $$\sqrt{2}d$$, that's why there is an extra 2 in the denominator in $$\frac{1}{2\sqrt{2}d^2}$$

cepheid
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In the picture I had $$\sqrt{2}d$$, that's why there is an extra 2 in the denominator in $$\frac{1}{2\sqrt{2}d^2}$$

flyingpig: Don't just guess. Read my previous post. You have a vector <-1, 1>, and you want to find a unit vector in the direction of that vector. Like I just said, in order to do that, you have to take this vector <-1,1>, and divide it by its magnitude. That suggests that you need to carry out these two steps.

1. Compute the magnitude of the vector <-1,1>.
2. Divide the vector <-1,1> by the number you found for the magnitude in step 1.

Can you carry out these two steps please? They will lead you to the right unit vector.

No, I am not guessing, I know what you are talking about. The unit vector for <-1,1> is

$$\frac{1}{\sqrt{2}}<-1,1>$$

But if I use that, it still wouldn't be right.

This is my reasoning, I'll write it like this

$$-k\frac{q^2}{\sqrt{2}(\sqrt{2d})^2}$$

$$\sqrt{2d}$$ This is the distance from 3 to 2 and that's why I had it in the denominator The other square root of 2 comes from taking the unit vector like you suggested.

Afterwards, I added then and I still get (if I get rid of the 2)

$$k\frac{q^2}{d^2}<\frac{1}{\sqrt{2}}, 2 - \frac{1}{\sqrt{2}}>$$

cepheid
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OH, sorry, yes, I do see what you mean. The force of 2 on 3 is given by

$$-k\frac{q^2}{2d^2}<-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}>$$

Now we can take the extra step to leave only kq^2/d^2 on the outside, and put everything else inside the angle brackets. Then the expression becomes:

$$k\frac{q^2}{d^2}<\frac{1}{2\sqrt{2}},-\frac{1}{2\sqrt{2}}>$$

So I agree with you after all. Now you have to add this to the force of charge 1 on charge 3, which can be re-written as:

$$k\frac{q^2}{d^2}<0,2>$$

where all I have done here is to take the factor of 2 in the numerator and distribute it to both terms inside the angle brackets.

Now, if you add these two forces together, you will get the correct answer given in your original post.

Yes, last night it came to me! Thanks