Electric Field Vector for point charge

[hteezy]

b]1. Homework Statement [/b]

Point charge q1= -5.00 is at the origin and point charge q2= +3.00 is on the -x axis at x= 3.00 cm. Point P is on the y-axis at y= 4.00 cm .

Calculate the electric fields E1 and E2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors (see example 21.6 in the textbook).

Express your answer in terms of the unit vectors \hat{i} , \hat{j}. E

Homework Equations

r = \sqrt{x^2 + y^2}

\hat{r} = (\vec{r}/r =x\hat{i} + y\hat{j}) / r
\vec{E} = k * q/r^2 8 \hat{r}

The Attempt at a Solution

So E1 will be the vector from q1 to point P
r = \sqrt{0^2 + .04^2} = .04 m

\hat{r} = (0\hat{i} + .04\hat{j})/.04 = 0\hat{i} + 1\hat{j}

E1 = (8.98 e 9) * (-5.00 e -9) / .04^2 (0\hat{i} + 1\hat{j})
= 0\hat{i} - 2.81 e 4\hat{j}

i did the exact same steps to find E2 and
E2 = 6.47 e 2\hat{i} + 8.62 e 2\hat{j}

am i doing something wrong accordin to the steps i used because my answers are wrong. and i even followed the way the steps were in the book.
i don't get it

 

[SammyS]

. . .
am i doing something wrong accordin to the steps i used because my answers are wrong. and i even followed the way the steps were in the book.
i don't get it

First of all, the LaTeX coding is all scrambled with plain text.

Here's a somewhat repaired version of what was posted:

Homework Statement
Point charge q₁= -5.00 is at the origin and point charge q₂= +3.00 is on the x-axis at x= 3.00 cm. Point P is on the y-axis at y= 4.00 cm .

Calculate the electric fields E₁ and E₂ at point P due to the charges q₁ and q₂. Express your results in terms of unit vectors (see example 21.6 in the textbook).

Express your answer in terms of the unit vectors $\hat{\imath} , ~\hat{\jmath}$.

E

Homework Equations

$\displaystyle r =\sqrt{x^2 + y^2 ~}$

$\displaystyle \hat{r} = \frac{\vec{r} } {r} =\frac{(x\,\hat{\imath}+ y\,\hat{\jmath})} {r}$

$\displaystyle\vec{E} = k \frac {q}{r^2} ~ \hat{r}$

Attempt at a Solution

So E₁ will be the vector from q₁ to point P
$r = \sqrt{0^2 + .04^2 ~} = .04$ m

$\displaystyle \hat{r} = \frac{(0\,\hat{\imath}+ .04\hat{\jmath})}{.04} = 0\hat{\imath} + 1\hat{\jmath}$

$\displaystyle \vec{E}_1= (8.98 \text{ e 9}) \frac{ -5.00 \text{ e -9}} {.04^2} (0~\hat{\imath} + 1~\hat{\jmath})$
= 0~\hat{\imath} - 2.81 \text{ e 4}~\hat{\jmath}
i did the exact same steps to find E₂ and

E₂ = $6.47 e 2 ~\hat{\imath} + 8.62 e 2~\hat{\jmath}$

am i doing something wrong according to the steps i used because my answers are wrong. and I even followed the way the steps were in the book.
I don't get it.