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.