Calculating Electric Field at a Point from Multiple Charges

[yevi]

I want to find an electric field in a point (x,y,z) generated by 2 charges:
q1 = q, and q2=2q, q1’s position = (1,2,3) and q2’s position = (4,5,6),

I tried to solve it by using superposition principle, adding the E of each charge to point x,y,z, I am doing something wrong probably with vectors calculation.

Please explain.

 

[G01]

Where in the calculation are you getting stuck? I'm not going to be able to give you any decent help if you don't show some work.

 

[yevi]

well,
E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1 + q/r2^2 * r2)
r1 and r2 vectors

 

[yevi]

r1=(x-1,y-2,z-3)
r2=(x-4,y-5,z-6)

after that and calculating all constants i get stuck.

 

[learningphysics]

well,
E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1 + q/r2^2 * r2)
r1 and r2 vectors

The vectors should be unit vectors...

 

[yevi]

yes unit vector,
I get: r1/|r1| but this doesn’t give me anything

 

[rootX]

If it should be like this:

E[x] = E1*(1/sqrt(14),0,0)+E2..

?
Thanks.

 

[learningphysics]

Yeah, how is the answer supposed to look? How do you know your answer is wrong?

 

[yevi]

I don't know the answer, but my answer has form with a lot of x,y and z variables.

 

[learningphysics]

well,
E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1 + q/r2^2 * r2)
r1 and r2 vectors

There should be 2q in the numerator or r2^2.

 

[rootX]

well,
E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1/|r1| + q/r2^2 * r2/|r2|)
r1 and r2 vectors

did you solve this one?
You would get lots of x,y,z vars, if you do

yea, changing to q2=2q should give the answer..

 

[yevi]

Yes, 2q.

I have a follow up question, to find field in (10,10,10) where q1=10^-6 q2=2*10^-6
and the answer for that is 10^-9K(21.09x’+17.76y’+14.43z’)

My answer and the point (10,10,10) doesn’t give this result.

 

[yevi]

rootX,
yes I get a mess of x,y and z.

 

[yevi]

I am doing something wrong with vectors, it's shouldn't be complicated...

 

[learningphysics]

Just leave it as the sum of the two fields... also you can factor out $$\frac{q}{({r_1^2})^{3/2}}$$ for the first part... and $$\frac{q}{({r_2^2})^{3/2}}$$ for the second part...

so all the messy stuff in the denominator can get factored out of the vectors... don't actually multiply out (x-a)^2 + (y-b)^2... etc.

 

[rootX]

I get
kq (0.123,1.17798..)
I guess I also messed up something..

 

[rootX]

ok, I am 55% sure that this answer is wrong: 10^-9K(21.09x’+17.76y’+14.43z’)

here's what I got from maxima:
kq(0.14227258171877 , 0.20662308551546 , 0.270973589312)

 

[learningphysics]

I'm getting the answer that is given: 10^-9K(21.09x’+17.76y’+14.43z’)

 

[yevi]

learningphysics,
can you please give your solution?

 

[learningphysics]

learningphysics,
can you please give your solution?

Basically I got:

$$kq*[\frac{(9,8,7)}{{194}^{3/2}} + 2*\frac{(6,5,4)}{{77}^{3/2}}]$$

this evaluates to the given answer.

 

[yevi]

lol
I was doing [sqrt(194)]^3/2 and [sqrt(77)]^3/2
Thanks

 

[rootX]

and I was using (1,2,3) for (9,8,4) and (4,5,6) for..

 

[yevi]

Another related question:

There are 2 linear charges distributed on y and x axis, segments lengths 0<=x<=l and 0<=y<=l.
Density of the charges is not uniformed: gamma(x)=bx and gamma(y)=by.
Need to fine E(0,0,z)

I tried to do following:

Because those 2 segments lay on axis x and axis y the needed Electric field is on axis z.
And because of symmetric I can calculate the contribution of one segments and contribution of the second will be the same.

So I need to find dE and then dE_z.

dE= $$\frac{kdq}{r^2}$$

dE_z= $$\frac{kdq}{r^2}$$ sin ($$\alpha$$)

r^2 = $$\sqrt{z^2+l^2}$$

Am I doing right so far?

 

[learningphysics]

You meant:

$$r=\sqrt{z^2+l^2}$$

Yes, what you've described will give you the field for one segment at the point z along the z-axis... then you can double it to get the field along the z-axis...

But you also need another component of the field at (0,0,z)

 

[yevi]

What do you mean by another component?

 

[yevi]

another question how do I convert the dq?

Is it bl?

 

[learningphysics]

another question how do I convert the dq?

Is it bl?

dq would be bxdx or bydy

 

[learningphysics]

What do you mean by another component?

There is a component that's parallel to the line y=x...

If the segments extend to the negative axes this component wouldn't be there... but since they only go from 0 to L... not -L to L... another component exists...

Imagine if we were dealing with equal point charges... one located at (L,0,0)... another at (0,L,0)... and you wanted the field at (0,0,0)... see the direction of the field? that's the direction of this extra component.

 

[yevi]

thanks

 

[yevi]

Hmm.

The given answer for this is: E_z=2kb($$\frac{z}{|z|}$$-$$\frac{z}{\sqrt{z^2+l^2}}$$)

But I don't get it right!

My integral from above is

$$\int\frac{2kbx}{z^2+l^2}sin(\alpha)dx$$

where I substitute sin($$\alpha$$)=$$\frac{z}{\sqrt{z^2+l^2}}$$