Calculating magnitude of total force (N)

[Mary O'Donovzn]

Homework Statement

You are given 3 charges (q1,q2,q3)

q1 has a magnitude (charge) of 7, q2 has a magnitude (charge) of -5, q3 has a magnitude (charge) of 3.

q1 has 2d co ordinates (3, 4), q2 has a 2d co ordinates (5, 9), q3 has a 2d co ordinates (7, 8).

For q1, q2 and q3 calculate the total force (N) and the angle it makes with the x-axis (radians) {there should be two}

Homework Equations

In your calculations you must use the following equation :

F =[K0 (q1q2)/d2 ] x r hat

Where K0 is Coulombs Constant = 8.9875 x 109 Nm2/C2

Also use the distance formula to get d the distance between the two

The Attempt at a Solution

[/B]
I couldn't even attempt this question without these things.

The real problem is r hat I don’t know how to calculate it and people just keep saying it’s a unit vector but I don’t understand how you use that.

Also do you add all three forces together like

F12, F13, F23 Or is there more than that?

 

[vela]

Homework Statement

You are given 3 charges (q1,q2,q3)

q1 has a magnitude (charge) of 7, q2 has a magnitude (charge) of -5, q3 has a magnitude (charge) of 3.

q1 has 2d co ordinates (3, 4), q2 has a 2d co ordinates (5, 9), q3 has a 2d co ordinates (7, 8).

For q1, q2 and q3 calculate the total force (N) and the angle it makes with the x-axis (radians) {there should be two}

Homework Equations

In your calculations you must use the following equation :

F =[K0 (q1q2)/d2 ] x r hat

Where K0 is Coulombs Constant = 8.9875 x 109 Nm2/C2

Also use the distance formula to get d the distance between the two

The Attempt at a Solution

[/B]
I couldn't even attempt this question without these things.

The real problem is r hat I don’t know how to calculate it and people just keep saying it’s a unit vector but I don’t understand how you use that.

Also do you add all three forces together like

F12, F13, F23 Or is there more than that?

Say you have two charges $q_1 = q_2 = +1.0\text{ C}$. The first charge is at the origin, and the second charge is at the point (3,4). Draw a picture of the two charges, and draw in the force on the second charge. How would you describe the direction of that force in words? Can you describe that mathematically as well, i.e., slope, angle, etc.?

 

[Mary O'Donovzn]

Say you have two charges q1=q2=+1.0 Cq_1 = q_2 = +1.0\text{ C}. The first charge is at the origin, and the second charge is at the point (3,4). Draw a picture of the two charges, and draw in the force on the second charge. How would you describe the direction of that force in words? Can you describe that mathematically as well, i.e., slope, angle, etc.?

If its with the origin would it be a right angles triangle so could I say θ = (x2-x1)i(hat) + (y2-y1) j (hat) divided by square root of (x2-x1)^2 + (y2-y1)^2

or use pythagoras's theorem?

 

[vela]

That's right. In this case, you'd have $\hat{r} = \left(\frac 35, \frac 45\right)$. The vector points away from the origin, where charge 1 is at, because the charges have the same sign and therefore repel. If the charges had opposite sign, you'd have $\hat{r} = -\left(\frac 35, \frac 45\right)$.

So what do you get for $\hat{r}$ for the original problem for the force on charge 2 due to charge 1?

 

[Mary O'Donovzn]

why are they both over 5?

I got (3/7) and (4/8)

I'm guessing that is very wrong

 

[vela]

In the example I gave, the vector $\vec{r}$ that goes from the first charge, which is at the origin, to the second charge, which is at (3,4) is $\vec{r} = 3\hat{i} + 4\hat{j}$. The 5 comes from length of this vector, which is given by $\| \vec{r} \| = \sqrt{3^2+4^2}$. You divide by that factor because you want $\hat{r} = \vec{r}/\|\vec{r}\|$ to have unit length:
$$\hat{r} = \frac 15 (3\hat{i} + 4\hat{j}) = \frac 35 \hat{i} + \frac 45 \hat{j}.$$ And if you were to calculate the length of $\hat{r}$, you'd find it indeed does have unit length:
$$\| \hat{r} \| = \sqrt{\left(\frac 35\right)^2 + \left(\frac 45\right)^2} = \sqrt{\frac{9+16}{25}} = 1.$$

In your problem, you have $(x_1, y_1) = (3,4)$ and $(x_2,y_2) = (5,9)$. The vector $\vec{r}$ is given by $\vec{r} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} = 2\hat{i} + 5 \hat{j}$. Can you take the rest from here?

 

[Mary O'Donovzn]

That is really clear example I think I got this :D
Thanks a million