Electric fields by point charges

[tkim90]

Homework Statement

*ignore the given numbers in the image, mine is different*

My numbers are:
Q1: 1nC
Q2: 8nC
Q3: -6nC

Homework Equations

$$E = \frac{kq}{r^2} * \hat{r}$$

The Attempt at a Solution

I only did (a1) so far. I know (a2) involves the exact same process (find r and plug it into the formula), except I don't know how to express it in coordinates since they're not on the same axes.

r = <? , ? , ?>
If i know r I think I could solve the problem.

thanks

 

[BigJDubs]

in advance.a1:Q1=1nC, Q2=8nC, r= <1, 2, 3>Electric Field:E = \frac{kq}{r^2} * \hat{r}E = \frac{1*9E9 * 1nC}{3^2} * \frac{<1,2,3>}{\sqrt{14}}E = \frac{3E9 * 1nC}{14} * \frac{<1,2,3>}{\sqrt{14}}E = \frac{3E9 * 1nC}{14} * <\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}>E = (\frac{3E9 * 1nC * 1}{14\sqrt{14}}, \frac{3E9 * 1nC * 2}{14\sqrt{14}}, \frac{3E9 * 1nC * 3}{14\sqrt{14}})

 

[kerimek]

Hello, it seems like you have a good understanding of the concept of electric fields by point charges. For part (a1), you correctly identified the equation for electric fields and found the distance, r, between the two charges. To solve part (a2), you can use the same process of finding the distance between the two charges, but instead of using coordinates, you can use the Pythagorean theorem to find the hypotenuse of the right triangle formed by the two charges. This will give you the value of r that you can then plug into the equation for electric fields. I hope this helps, and keep up the good work with your problem-solving skills!