# I Radial Vector in Cartesian form

#### Arman777

Gold Member
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

Homework Helper
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 $\sqrt {x^2+y^2}>0$.

• Arman777 and PeroK

#### PeroK

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Gold Member
2018 Award
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

Gold Member
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

Gold Member
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

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

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#### PeroK

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2018 Award
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

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Gold Member
2018 Award
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

Gold Member
It looks like you have $\vec d = (d, d)$ there.
my mistake $\vec{d} = d\hat{i}$