Why does the unit vector r-hat always point away from a charge?

[negation]

In my textbook, it states that given q1 and q2 where both charges are positive and of equal magnitude, the unit vector r-hat always points away from q1. Why is this so?
What if both charges are of opposite charge? Does r-hat still points away?

 

[Doc Al]

In my textbook, it states that given q1 and q2 where both charges are positive and of equal magnitude, the unit vector r-hat always points away from q1. Why is this so?
What if both charges are of opposite charge? Does r-hat still points away?

Presumably the unit vector is describing the field from q1, which is outward. (The field from a negative charge would point inward.)

Do you have a diagram of what they are describing?

 

[ZapperZ]

In my textbook, it states that given q1 and q2 where both charges are positive and of equal magnitude, the unit vector r-hat always points away from q1. Why is this so?
What if both charges are of opposite charge? Does r-hat still points away?

Er.. it doesn't have to! It depends on where you put the origin of your coordinate axis. If you put the origin right in the middle in between the two charges, then the radial unit vector points AT each of the charge when you are in between the two charges, and then away from the charges if you are beyond the two charges.

The radial unit vector is a function of where you define the origin of your coordinate system. It has nothing to do with being tied to a particular charge. In other words, it is math. However, placing it in an appropriate location can simplify the mathematics. That is why it is usually placed at the location of the source charge!

Zz.

 

[Chestermiller]

In my textbook, it states that given q1 and q2 where both charges are positive and of equal magnitude, the unit vector r-hat always points away from q1. Why is this so?
What if both charges are of opposite charge? Does r-hat still points away?

The unit vector r-hat is a geometric quantity, not a physical quantity. If one of the charges is at the origin and you are employing spherical coordinates, in spherical coordinates the radial unit vector r-hat points away from the origin.

Chet

 

[negation]

Presumably the unit vector is describing the field from q1, which is outward. (The field from a negative charge would point inward.)

Do you have a diagram of what they are describing?

Yes I do. But how do I upload it? I do not have a scanner so the only option is to utilize my phone's camera.

 

[negation]

The unit vector r-hat is a geometric quantity, not a physical quantity. If one of the charges is at the origin and you are employing spherical coordinates, in spherical coordinates the radial unit vector r-hat points away from the origin.

Chet

Isn't it the case that if we were measuring the electric force at a point at <5,5> then r-hat would be <5,5>?-that is, r-hat is a vector to provide the coordinate of the point at which we are measuring an electric force?
If the point at which we are measuring the electric force is at <0,0>, where would r-hat point to?

 

[ZapperZ]

Isn't it the case that if we were measuring the electric force at a point at <5,5> then r-hat would be <5,5>?-that is, r-hat is a vector to provide the coordinate of the point at which we are measuring an electric force?

Move your origin to (-10, 0). What is "r-hat" now?

Zz.

 

[craigi]

Yes I do. But how do I upload it? I do not have a scanner so the only option is to utilize my phone's camera.

View attachment 68552

It's just a notational convenience to ensure that the formula always gives the correct direction.

It's a unit vector that points away from the other charge.

 

[negation]

Move your origin to (-10, 0). What is "r-hat" now?

Zz.

<-5,5>?

 

[ZapperZ]

<-5,5>?

I think you have a problem with geometry, not with physics.

Zz.

 

[Chetlin]

Isn't it the case that if we were measuring the electric force at a point at <5,5> then r-hat would be <5,5>?

It's a unit vector, so it's magnitude has to be 1, so you have to normalize it too by dividing each component by its magnitude. You'd get \hat{r} = \left\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle. The purpose of a unit vector is to provide the direction of something, so you can multiply it by the magnitude of something else (in this case, the electric force) to get the force vector with the correct magnitude and direction. (You're multiplying the magnitude of the force by the magnitude of the unit vector, which is 1, so the resulting vector has the same magnitude as that of the force, but also in the same direction as the unit vector.)

 

[negation]

I think you have a problem with geometry, not with physics.

Zz.

If the coordinate shifts by -10, am I not suppose to move the x-coordinate by that amount?

 

[craigi]

<-5,5>?

Check that you know how to take the magnitude of a vector and what a unit vector is.

 

[negation]

Check that you know how to take the magnitude of a vector and what a unit vector is.

The magnitude of a vector is sqrt(x^2 + y^2+z^2) and a unit vector i-hat represents one unit of a component in the x-direction.

 

[ZapperZ]

This thread is moved to the mathematics forum from the General Physics forum, because this is no longer physics.

Zz.

 

[craigi]

The magnitude of a vector is sqrt(x^2 + y^2+z^2) and a unit vector i-hat represents one unit of a component in the x-direction.

The first part is correct. Check the second part again.

 

[Doc Al]

Yes I do. But how do I upload it? I do not have a scanner so the only option is to utilize my phone's camera.

View attachment 68552

Ah, now I see what's going on. Chestermiller is right.

When using Coulomb's law in vector form, that unit vector always points outward from q1. Then you can use the signs of q1 and q2 to determine the direction of the force that q1 exerts on q2. When they have the same sign, the force acts in the direction of the unit vector.

The unit vector just describes the direction from q1 to q2.

 

[negation]

The first part is correct. Check the second part again.

Isn't i-hat one unit component of x?

The mathematical definition is e₁ or vector r / magnitude r

 

[negation]

Ah, now I see what's going on. Chestermiller is right.

When using Coulomb's law in vector form, that unit vector always points outward from q1. Then you can use the signs of q1 and q2 to determine the direction of the force that q1 exerts on q2. When they have the same sign, the force acts in the direction of the unit vector.

The unit vector just describes the direction from q1 to q2.

and if the force were from q2 to q1? the unit vector would point to the left?

In the second diagram however, the electric force is from q2 to q1 yet the unit vector points to the right of q2.

 

[craigi]

Isn't i-hat one unit component of x?

The mathematical definition is e₁ or vector r / magnitude r

You're close, but this isn't quite correct. Where are you learning this from?

 

[Doc Al]

and if the force were from q2 to q1? the unit vector would point to the left?

No, the unit vector describes the direction from q1 to q2. The force may be in the same direction or in the opposite direction, depending on the signs of the charges.

In the second diagram however, the electric force is from q2 to q1 yet the unit vector points to the right of q2.

That's right.