Undergrad Radial Vector in Cartesian form

[Arman777]

If I wanted to write $\hat{r}$in terms of $\hat{x}$and $\hat{y}$, is it $\frac{\hat{x} + \hat{y}}{\sqrt{2}}$ ?

 

[tnich]

If I wanted to write $\hat{r}$in terms of $\hat{x}$and $\hat{y}$, is it $\frac{\hat{x} + \hat{y}}{\sqrt{2}}$ ?

Only if $x=y\ge 0$.
In general $\hat r= \frac{x\hat x + y\hat y} {\sqrt {x^2+y^2}}$ for $\sqrt {x^2+y^2}>0$.

 

[PeroK]

Only if $x=y\ge 0$.
In general $\hat r= \frac{x\hat x + y\hat y} {\sqrt {x^2+y^2}}$ for $\sqrt {x^2+y^2}>0$.

... as

1) $\vec r= x\hat x + y\hat y$

2) $\hat r = \frac{\vec r}{|\vec r|}$

3) $|\vec r| = \sqrt{x^2 + y^2}$

 

[Arman777]

Thanks

 

[Stephen Tashi]

If I wanted to write $\hat{r}$in terms of $\hat{x}$and $\hat{y}$, is it $\frac{\hat{x} + \hat{y}}{\sqrt{2}}$ ?

What does the "$\hat{}$" in your notation signify? Is it only to indicate that variables are vectors? - or does it indicate vectors of length 1?

What is your definition of $\hat{r}$?

 

[Arman777]

I found an E field in the form of $\vec{E} = C(\frac{1} {|\vec{r}|} - \frac{1} {|\vec{r} - \vec{d}|})\hat{r}$ where C is a constant.

I need to transform this into x,y coordinates. So I wrote

$\vec{E} = C(\frac{1} {\sqrt{x^2 + y^2}} - \frac{1} {\sqrt{(x-d)^2 + (y)^2)}}) \frac{x\hat{i} + y\hat{j}}{\sqrt{x^2 + y^2}}$

 

[PeroK]

I found an E field in the form of $\vec{E} = C(\frac{1} {|\vec{r}|} - \frac{1} {|\vec{r} - \vec{d}|})\hat{r}$ where C is a constant.

I need to transform this into x,y coordinates. So I wrote

$\vec{E} = C(\frac{1} {\sqrt{x^2 + y^2}} - \frac{1} {\sqrt{(x-d)^2 + (y-d)^2)}}) \frac{x\hat{i} + y\hat{j}}{\sqrt{x^2 + y^2}}$

It looks like you have $\vec d = (d, d)$ there.

 

[PeroK]

What does the "$\hat{}$" in your notation signify? Is it only to indicate that variables are vectors? - or does it indicate vectors of length 1?

What is your definition of $\hat{r}$?

$\hat r$ is a unit vector.

 

[Arman777]

It looks like you have $\vec d = (d, d)$ there.

my mistake $\vec{d} = d\hat{i}$