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

[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.

 

[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.
$$V(\vec{R})=\frac{-kq}{|\vec{R}|}$$
$$\vec{E}(\vec{R})=\frac{kq\vec{R}}{|\vec{R}|^3}$$

 

[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.

 

[Sean1218]

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.
$$V(\vec{R})=\frac{-kq}{|\vec{R}|}$$
$$\vec{E}(\vec{R})=\frac{-kq\vec{R}}{|\vec{R}|^3}$$

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

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.

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

 

[SammyS]

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.

The answer to your question is implied by the answer to the question asked by the title of this thread:

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.

 

[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.

 

[Sean1218]

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.

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).

 

[SammyS]

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).

Please show your work for this.

 

[Sean1218]

Please show your work for this.

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?

 

[Steely Dan]

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?

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.