# Can you not separate a scalar into x and y components?

1. Apr 18, 2012

### Sean1218

A scalar like electric potential.

Say I have a positive charge, and 4m to the right, and 3m up is a point P.

If I wanted to calculate the potential at point P, I'd use V=kQ/r (r=√(4^2 + 3^2)).

But I'm confused about why finding the potential at 4m to the right (the x component), and the potential at 3m above the charge (y component) and using c=√(x^2 + y^2) wouldn't give the same answer.

And I realize that this isn't independent for scalars, because doing this to find electric field would be wrong as well.

Last edited: Apr 18, 2012
2. Apr 18, 2012

### Mindscrape

What is c? The scalar potential is itself a function of a vector if that clears up anything. Similarly, the electric field is a vector that itself is a function of a vector. i.e.
$$V(\vec{R})=\frac{-kq}{|\vec{R}|}$$
$$\vec{E}(\vec{R})=\frac{kq\vec{R}}{|\vec{R}|^3}$$

Last edited: Apr 18, 2012
3. Apr 18, 2012

### Steely Dan

Well, if the electrostatic potential could have "components," then it stands to reason that any arbitrary function could. Take $z = x^2 + y$, for example. It is clearly true that $z(4,3) = 19$, but $z(4,0) = 16$ and $z(0,3) = 3$, so $z(4,3) \neq (z(4,0)^2 + z(0,3)^2)^{1/2}$. Ordinary functions of space simply do not work the same way as vectors. So in a sense, it's meaningless to compare scalar quantities and vector quantities this way, because vectors are geometrical quantities which are specifically constructed to satisfy the property that you can use the Pythagorean theorem to determine their magnitude.

4. Apr 18, 2012

### Sean1218

Sorry, c is just the hypotenuse so electric potential at P (or electric field at P).

Then why doesn't this method work with electric field, as it's a vector?

5. Apr 18, 2012

### SammyS

Staff Emeritus
Can you not separate a scalar into x and y components?
The answer is that a scalar cannot be separated into x & y components, because a scalar is not a vector quantity. It may have a magnitude, but it does not have a direction associated with it.

6. Apr 18, 2012

### Mindscrape

So, what are you saying doesn't work? The distance to the point charge is the hypotenuse. The hypotenuse itself does have an x component and a y component. The electric potential, however, does not have any components because it just depends on the hypotenuse. The electric field, on the other hand, not only depends on the hypotenuse, but where the hypotenuse is located.

7. Apr 18, 2012

### Sean1218

I understand now regarding electric potential and other scalars, thanks everyone.

What doesn't work is finding the x and y components of the electric field and using pythagorean theorem to find the electric field at that point. This method should give an identical answer to using E=kq/r2 where r=5 (distance between P and q), but it doesn't (for me at least).

8. Apr 18, 2012

### SammyS

Staff Emeritus

9. Apr 18, 2012

### Sean1218

With a charge of 8E-6 C

x: kq/r2 = (9E9)(8E-6)/(42) = 4500

y: kq/r2 = (9E9)(8E-6)/(32) = 8000

E=sqrt(45002 + 80002)
E=9179

or:

r=5
kq/r2 = (9E9)(8E-6)/(52) = 2880 (the correct answer)

Am I just doing the first part wrong?

10. Apr 19, 2012

### Steely Dan

Right, this is not correct. The x component of the electric field vector at point P is not the same thing as the value of the electric field at x = 4, y = 0. Those are two unrelated quantities. The x component of E at point P is the magnitude of E, multiplied by the cosine of the angle between the vector and the x-axis.