# Calculating electric field

This problem is giving me a lot of problems. So $E=k*q/d^2$. We'd want to find the distance from q1 to P, which is .1 meters (not cm) using pythagorean thereom. So we know k, which is just $9x10^9$ times q1 which is $-2.4u$ where $u=10^{-6}$ divided by $r^2$ which is just $.1^2$. Then I get $-216,000,000$. To get x component, I take that number multiplied by $cos45$ to come up with a final answer of $-108,000,000$. And for point q2, I dont think there is an x component for the electric field since its right below the point P.

What have I done wrong?

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I suggest you use vector maths to simplify things here. –  Noldorin Jan 24 '11 at 1:46
What do you mean? –  maq Jan 24 '11 at 1:48
Well it's a 2-dimensional problem, so it is much more convenient to use vectors and their corresponding notation here than treat the dimensions separately. –  Noldorin Jan 24 '11 at 1:53
I'm surprised that any physics course would not explain vector algebra before teaching E&M... but anyways, why are you using Cos(60) to calculate the x component? –  stl_piznaul Jan 24 '11 at 2:01
A gneral comment I make to my students when I see solution attempts like this is "You're using numbers too soon." Keep your solution in symbols as long as you can, and you notice that you can factor out the $k$ which will spare you all the zeros right up until the end. Secondly, as others have noted there does not appear to be a 60 degree angle anywhere in the problem. –  dmckee Jan 24 '11 at 2:17

The electric field is the vector sum $${\vec E}~=~\frac{1}{4\pi\epsilon_0}\Big(\frac{q_1{\bf n}_1}{2d^2}~+~\frac{q_2{\bf n}_2}{d^2}\Big)$$ so the components of the field are $$E_x~=~\frac{1}{4\pi\epsilon_0}\frac{q_1}{2\sqrt{2}d^2}$$ $$E_y~=~\frac{1}{4\pi\epsilon_0}\frac{q_1}{d^2}\frac{1~+~2\sqrt{2}}{2\sqrt{2}}$$ The rest is plug and grind on numbers.

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Ok so I came up with an answer of -15585 but the system told me I was off by a power 10, so I added two more zeros to get the right answer. But I have no idea what I did or what you did :( What is n1? I put q1 has -2.4 then multiplied by $10^{-6}$. Then divided by $2*.7^2$ which is .98. Mutliplied by K, and took the cos45. What am I still doing wrong? I got the right answer, but I just want to understand it more.. –  maq Jan 24 '11 at 2:38
@mohabitar: $\mathbf n_1$ is a unit vector. That means it is an arrow with unit length. E.g. $(1, 0)$ would be an arrow pointing right and $(0, 1)$ would be an arrow pointing up. For general arrow you'd get a $(x, y)$ but you have to normalize it so that it only contains direction information. So e.g. here ${\mathbf n_1} = {1 \over \sqrt 2} (1, 1)$ for a north-east arrow. Electromagnetism obeys the principle of superposition and the fields are generated by charges sitting at some points of space. You can represent these points by vectors and then you just add the fields to obtain the result. –  Marek Jan 24 '11 at 9:36

Rather than answer the question, I assume you just started a physics course (my kids are in their first week this semester). This course should have had vector algebra, and probably other math as a prerequisite. If you don't have the prerequisites, then you have two choices, (1) drop the course and take it later after satisfying the prereqs, or (2) try to tough it out, which means you will have to rapidly pick up the missing prereqs. If you are a really good student, and the gaps aren't too great, I'd say go for it, otherwise option (1) would be your better choice. Note, option two may mean you have doubled or trippled the amount of work needed to pas the course, and you'll need to be an auto-didact (self learner). You can probably find free course notes on opencourseware, download the appropriate ones and see if you think you are up to the task.

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Well I'm in Calc III now, so we just started Vector stuff, but we had a basic vector review in last semester's Physics course. I just need a little review maybe.. –  maq Jan 24 '11 at 6:19

First off, I don't want to answer the question for you, but your distance between q1 and point P seems to be incorrect.

You should really be working with units as well - this will help you catch any mistakes you may be making.

Your general reasoning seems to be on track though.

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How is it incorrect? –  maq Jan 24 '11 at 1:51
Oh .01 is in meters. I converted from cm to m because thats the way the E equation is set up.. –  maq Jan 24 '11 at 1:53