- #1

Zack K

- 166

- 6

## Homework Statement

A charged particle has an electric field at ##\langle -0.13, 0.14, 0 \rangle## m is ##\langle 6.48\times10^3, -8.64\times10^3, 0 \rangle## N/C. The charged particle is -3nC. Where is the particle located?

## Homework Equations

##\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|} \hat {r}##

## The Attempt at a Solution

The solution was already given since this was a textbook example. Going through the process:

##|\vec E|=\sqrt{(6.4\times 10^3)^2 + (-8.64\times 10^3)^2} N/C = 1.08\times 10^4 N/C##

They then present the modified equation: ##|\vec E|=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|}##

My question is why is that equation true? Why does substituting the vector ##\vec E## for the magnitude ##|\vec E|## get rid of the unit vector ##\hat {r}##? The book never explained if they are related.

Edit: I didn't write down the entire solution since those steps make sense to me, it's only this part.

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