Creating an Electrical field equation

[Alv95]

Homework Statement

I have to find a general equation to find the magnitude of the resultant electrical field vector on a positive test charge by two charges of the same magnitude but opposite signs.
The two charges are equidistant from the y-axis and are on the x-axis.
Q1= positive charge
Q2= negative charge
a= distance between a charge and the y-axis
(x;y)= coordinates of the test charge
k ( constant) = 899 000 000

Homework Equations

All electromagnetism related equations

The Attempt at a Solution

I found the distance r1 between Q1 and the test charge:
$$\sqrt{(x+a)^2+y^2}$$
I found the distance r2 between Q2 and the test charge:
$$\sqrt{(x-a)^2+y^2}$$
I found the magnitude of E1:
$$\frac{kQ}{r1^2}= \frac{kQ}{(x+a)^2+y^2}$$
I found the magnitude of E2:
$$\frac{kQ}{r2^2}= \frac{kQ}{(x-a)^2+y^2}$$
I found the angle β between E1 and the x-axis:
$$\arctan{(\frac{y}{x+a})}$$
I found the angle γ between E2 and the x-axis:
$$\arctan{(\frac{y}{x-a})}$$
I found the x component of E1:
$$E1\cos{β}= \frac{kQ}{(x+a)^2+y^2}\cos(\arctan{(\frac{y}{x+a})})$$
I found the x component of E2:
$$E1\cos{γ}= \frac{kQ}{(x-a)^2+y^2}\cos(\arctan{(\frac{y}{x-a})})$$
I found the y component of E1:
$$E1\sin{β}= \frac{kQ}{(x+a)^2+y^2}\sin(\arctan{(\frac{y}{x+a})})$$
I found the y component of E2:
$$E1\sin{γ}= \frac{kQ}{(x-a)^2+y^2}\sin(\arctan{(\frac{y}{x-a})})$$

At this stage I thought of doing the sum of te x and y components and then use the pythagorean theorem but I'm not sure of the signs... Is it right? Do you have any Idea on how to do it?
I have to find an always valid equation :)

 

[tiny-tim]

Welcome to PF!

Hi Alv95! Welcome to PF!

(try using the X₂ button just above the Reply box )

At this stage I thought of doing the sum of te x and y components and then use the pythagorean theorem but I'm not sure of the signs... Is it right?

The question asks for the magnitude, so yes, sum the components and use Pythagoras.

(btw, also use Pythagoras to find the cos and sin).

As to signs, one arrow will point towards one charge, and the other arrow will point away from the other charge …

if you draw a diagram, it should be obvious which components are negative. ​

 

[Alv95]

As to signs, one arrow will point towards one charge, and the other arrow will point away from the other charge …
if you draw a diagram, it should be obvious which components are negative. [/INDENT]

But if I change the position of the test charge I will have to sum some components or subtract other ones... :) I think I should incorporate in my equation a way to know the direction of E1 and E2 so that when I sum them it will automatically do the right operation depending on the position pf the test charge... But I don't know how :)

Thank you very much for your reply and for your help :)
Any suggestions?
I get: =
$$\sqrt{(E1\cos\beta \pm E2\cos\gamma)^2 +(E1\sin\beta \pm E2\sin\gamma)^2}$$
but I don't know which signs to use and I don't know if it will work :)

P.S. The teacher said that if x=0.55 and y=0.77 and a=0.5 and Q=2x10^-6 I should get Etot= 24760

 

[tiny-tim]

Hi Alv95!

… I don't know which signs to use and I don't know if it will work :)

you do the magnitude, and then the sin and cos

the magnitude is always positive, so that's not a problem!

then for the positive charge, the y component will always be y/hypotenuse,

and for the negative charge, the y component will always be -y/hypotenuse

(and something similar for the x components)

 

[Alv95]

The teacher said that I only have to find the formula for the magnitude of the resultant electrical field so I think it is not necessary to calculate the sin and cos. For what concerns the signs I know that the magnitude is always positive but I would like to know if the formula should be:
$$Etot=\sqrt{(E1\cos\beta+E2\cos\gamma)^2 + (E1\sin\beta+E2\sin\gamma)^2}$$ or
$$Etot=\sqrt{(E1\cos\beta-E2\cos\gamma)^2 + (E1\sin\beta-E2\sin\gamma)^2}$$
If the components of the component vectors automatically change their sign multiplicating by sin and cos it should be the first equation I think :)
Is it right? Any suggestions? Have you found any other equations?
Thank you very much ;)

 

[tiny-tim]

The teacher said that I only have to find the formula for the magnitude of the resultant electrical field so I think it is not necessary to calculate the sin and cos.

I'm confused …

your formula has cos and sin of β and γ …

and they're not given in the question, so surely you do need to find them?

 

[Alv95]

Sure :) I thought you were saying cos and sin of the total electrical field :)
Well I already know the ones that you say: I found that:
$$\cos\beta= \cos(\arctan(\frac{y}{x+a}))$$
$$\sin\beta= \sin(\arctan(\frac{y}{x+a}))$$
$$\cos\gamma= \cos(\arctan(\frac{y}{x-a}))$$
$$\sin\gamma= \sin(\arctan(\frac{y}{x-a}))$$

For instance the Etot equation that is written on my previous post can be written as ( extended form):

$$Etot = \sqrt{\{\frac{kQ}{(x+a)^2+y^2}\cos[\arctan(\frac{y}{x+a})]+\frac{kQ}{(x-a)^2+y^2}\cos[\arctan(\frac{y}{x-a})]\}^2+\{\frac{kQ}{(x+a)^2+y^2}\sin[\arctan(\frac{y}{x+a})]+\frac{kQ}{(x-a)^2+y^2}\sin[\arctan(\frac{y}{x-a})]\}^2}$$ using only the data provided... Will it be right? ;)

 

[tiny-tim]

Well, eg
$$\cos\beta= \cos(\arctan(\frac{y}{x+a}))$$

= (x+a)/√((x+a)² + y²)

 

[Alv95]

= (x+a)/√((x+a)² + y²)

Wich one will work best? And for the $\pm$ in my total general equation (see below) is it + or is it - (inside the two squared brackets) ? :)

My total general equation can be rewritten as following, following your suggestion:

$$Etot =\sqrt{[\frac{kQ}{(x+a)^2+y^2}\frac{x+a}{\sqrt{(x+a)^2+y^2}}\pm\frac{kQ}{(x-a)^2+y^2}\frac{x-a}{\sqrt{(x-a)^2+y^2}}]^2+[\frac{kQ}{(x+a)^2+y^2}\frac{y}{\sqrt{(x+a)^2+y^2}}\pm\frac{kQ}{(x-a)^2+y^2}\frac{y}{\sqrt{(x-a)^2+y^2}}]^2}$$

Is this one better than the other exented one ? :)

 

[Alv95]

Eureka! :) I have eventually found the final total expanded equation. Thanks for your help tiny-tim! :D
It works with the data that the teacher has given me! I will try with other ones :)
$$Etot =\sqrt{[\frac{kQ}{(x+a)^2+y^2}\frac{x+a}{\sqrt{(x+a)^2+y^2 }}+\frac{-kQ}{(x-a)^2+y^2}\frac{x-a}{\sqrt{(x-a)^2+y^2}}]^2+[\frac{kQ}{(x+a)^2+y^2}\frac{y}{\sqrt{(x+a)^2+y^2}} +\frac{-kQ}{(x-a)^2+y^2}\frac{y}{\sqrt{(x-a)^2+y^2}}]^2}$$

Final form ( could someone check if it is right?) :) =

$$Etot =kQ\times\sqrt{\{\frac{x+a}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{x-a}{[(x-a)^2+y^2]^(\frac{3}{2})}\}^2+\{\frac{y}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{y}{[(x-a)^2+y^2]^(\frac{3}{2})}\}^2}$$

Even if the teacher has not asked to do so... is there a way to find the direction of the Etot vector? ;) ...

 

[Alv95]

Eureka! :) Thanks for your help tiny-tim! :D
Final form ( could someone check if it is right?) :) =

$$Etot =kQ\times\sqrt{\{\frac{x+a}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{x-a}{[(x-a)^2+y^2]^(\frac{3}{2})}\}^2+\{\frac{y}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{y}{[(x-a)^2+y^2]^(\frac{3}{2})}\}^2}$$

It is right! :)
For what concerns its direction I think to know how to do it but now I have to go to school :) I will post it later ;)

 

[tiny-tim]

A quick latex tip:

after ^ or _ , you need to put everything inside {} unless it's only one character (eg 2)

 

[Alv95]

The equation for the angle $ψ$ of the $Etot$ vector is:
$$\psi = \arctan(\frac{\frac{x+a}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{x-a}{[(x+-a)^2+y^2]^(\frac{3}{2})}}{\frac{y}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{y}{[(x-a)^2+y^2]^(\frac{3}{2})}})$$

I showed my equations yesterday in class (I was the only one who did the homework) but no one cared and as usual the other ones have not been punished by the teacher for not having done their homework :(

P.S. It was very funny aniway to create this formula :D

 

[tiny-tim]

… I was the only one who did the homework …

Well done!

btw, this is what the latex should look like (check the code by hitting the reply button)…
$$\psi = \arctan\left(\frac{\frac{x+a}{[(x+a)^2+y^2]^{\frac{3}{2}}}-\frac{x-a}{[(x+-a)^2+y^2]^{\frac{3}{2}}}}{\frac{y}{[(x+a)^2+y^2]^{\frac{3}{2}}}-\frac{y}{[(x-a)^2+y^2]^{\frac{3}{2}}}}\right)$$
and you can simplify it to …
$$\psi = \arctan\left(\frac{(x+a)[(x-a)^2+y^2]^{\frac{3}{2}} -(x-a)[(x+a)^2+y^2]^{\frac{3}{2}}}{y[(x^2+y^2+a^2)^2-4a^2x^2]^{\frac{3}{2}}}\right)$$

 

[Alv95]

Thank you very much tiny-tim for your precious Latex tips and your suggestions :)

Here is a little resumé of all this thread:

The resultant Electrical field vector ($Etot$) on a test charge due to two charges of equal magnitude but opposite sign, equidistant from the y-axis and lying on the x-axis, and the angle $ψ$ that it forms with the x-axis, are given by:

$$Etot =kQ\times\sqrt{\left\{\frac{x+a}{[(x+a)^2+y^2]^{\frac{3}{2}}}-\frac{x-a}{[(x-a)^2+y^2]^{\frac{3}{2}}}\right\}^2+\left\{\frac{y}{[(x+a)^2+y^2]^{\frac{3}{2}}}-\frac{y}{[(x-a)^2+y^2]^{\frac{3}{2}}}\right\}^2}$$

$$\psi = \arctan\left(\frac{(x+a)[(x-a)^2+y^2]^{\frac{3}{2}} -(x-a)[(x+a)^2+y^2]^{\frac{3}{2}}}{y[(x^2+y^2+a^2)^2-4a^2x^2]^{\frac{3}{2}}}\right)$$

Does anyone know if with my ψ equation we get the counter-clockwise angle as shown or the clockwise angle? :)