- #1

jonyamikun

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## Homework Statement

A -10.0 nC charge is located at position (x,y) = (2cm, 1cm).

At what (x,y) position(s) is the electric field:

Here is where they give me the different electric fields, one of which is: 161,000[tex]\hat{i}[/tex] - 80,500[tex]\hat{j}[/tex] N/C

The answers must be in cm.

## Homework Equations

[tex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}[/tex]

## The Attempt at a Solution

What I first did was plug in the the numbers I'm given:

E = 161000

epsilon_0 = 8.85x10^-12

q = 10x10^-9

Then I try to get r^2 by itself by multiplying it with both sides and then dividing both sides by 161000.

I now have: [tex]r^2 = \frac{10\times10^{-9}}{4\pi(8.85\times10^{-12})(161000)}[/tex]

Which gives me r^2 = 5.85*10^-4

Squarerooting the answer gives me 0.0236 meters which converted to cm is 2.36 rounded to 2.4.

Taking 2.4, I have added to the initial 2cm in the beginning of the problem to get 4.4 and have tried subtracting from 2 to get -0.4 but both are wrong. I did the first problem (E = -225,000 i-hat) the exact same way and I got the correct answer. My friend tried it out and also got the same answers as me.

http://i385.photobucket.com/albums/oo297/jonyamikun/2664.jpg

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