Question

A little question about the total strength and direction of electric field.

Answer 1

The distance from the bottom charge is: \(\sqrt{(2x)^2+(x)^2}=\displaystyle\sqrt{5}{\tiny~}x\)
(Well, in my problem \(x\) is positive, so no abs. value needed)

Answer 2

Then, the first point charge acts on the particle as follows:
(Conversion nC -> C, and cm -> m, are made ... you can note that here)

\(E_1=\displaystyle \frac{8.9875\times 10^9\times b\times 10^{-9}}{((1/100)x)^2}=\displaystyle \frac{8.9875\times 10^9\times b\times 10^{-9}}{10^{-4}x^2}=\frac{8.9875b\times 10^4}{x^2}N/C\)

Answer 3

N/C, at the end.

Answer 4

Am I doing it right so far?

Answer 5

\(E_2=\displaystyle \frac{8.9875\times 10^9\times b\times 10^{-9}}{((1/100)\sqrt{5}x)^2}==\frac{8.9875b\times 10^4}{5x^2}N/C\)

Answer 6

Now, I have to find their directions:

For charge 1 (the upper charge), it is to the left.
For charge 2, \(\theta=\arctan(2x/x=2)\approx63.43 ^\circ\).

Answer 7

yes, looks ok.
change each to a <x,y> vector with strength "diluted" by 1/r^2
at the point in question. then add the two vectors

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @idku
Now, I have to find their directions:

For charge 1 (the upper charge), it is to the left.
For charge 2, \(\theta=\arctan(2x/x=2)\approx63.43 ^\circ\).
\(\color{#0cbb34}{\text{End of Quote}}\)

in terms of your drawing, it is to the **right**

the E field presupposes a +ve test charge at the point in question

Answer 9

\(\displaystyle E_1=\langle\frac{8.9875b\times 10^4}{x^2},0\rangle \)

\(\displaystyle E_2=\langle\frac{8.9875b\times 10^4}{5x^2}\sin(\arctan 2),\frac{8.9875b\times 10^4}{5x^2}\cos(\arctan 2)\rangle \)

Answer 10

Calculation:
\(\displaystyle \sin(\arctan 2)=\frac{\tan(\arctan 2)}{\sqrt{\tan^2(\arctan 2)+1}}=\frac{2}{\sqrt{5}} \)


\(\displaystyle \cos(\arctan 2)=\frac{1}{\sqrt{\tan^2(\arctan 2)+1}}=\frac{1}{\sqrt{5}} \)

(this is positive (without \(\pm\)) since sin(arctan 2) and cos(arctan 2) are \(\ge\)0)

Answer 11

\(\displaystyle E_1=\langle\frac{8.9875b\times 10^4}{x^2},0\rangle\)

\(\displaystyle E_2=\langle\frac{(2)8.9875b\times 10^4}{5\sqrt{5}x^2},\frac{8.9875b\times 10^4}{5\sqrt{5}x^2}\rangle\)

Answer 12

Then, the resultant strengths in x and y positive directions are as follows:

\(\displaystyle E_y=0+\frac{8.9875b\times 10^4}{5\sqrt{5}x^2}\)

\(\displaystyle E_x=\left(\frac{2}{5\sqrt{5}}+1\right)\frac{8.9875b\times 10^4}{x^2}=\left(\frac{2+5\sqrt{5}}{5\sqrt{5}}\right)\frac{8.9875b\times 10^4}{x^2}\)

Answer 13

Then, the angle would be: \(\arctan (E_y/E_x)\)

and the total strength of the electric field should follow by the Pythagorean theorem:
\(E_{net}=\sqrt{[E_y]^2+[E_x]^2}\)