Electric field due to 2 charged spheres

In summary, Johnathan attempted to solve the homework by using the superposition principle and an approximation for the field outside the body. However, he ran into trouble with the third power in the norm of the vector and was not able to completely solve the problem.
  • #1
Yoni V
44
0

Homework Statement


The first part is to calculate the electric field everywhere in space given a body of 2 spheres of radius R and distance d apart (d<R), located on the z-axis, with charge density ρ and -ρ.
Of course when r>>R this is essentially a dipole.
The second part is to approximate the field outside the body given R>>d, i.e. the 2 spheres almost entirely overlap.

Homework Equations


E=E1+E2

The Attempt at a Solution


Using the superposition principle I got to the following expression for the electric field outside the body (before the approx.):
E=kρ4/3piR^3[(r-d/2z-hat)/|r-d/2z-hat|^3-(r+d/2z-hat)/|r+d/2z-hat|^3]
(sorry, couldn't get the latex to work...)

Now, if R>>d I approximated it to be zero:
E=kρ4/3piR^3[r/|r|^3-r/|r|^3]=0

It kinda makes sense as a crude approximation because the charges almost cancel entirely. But I'm pretty sure this not how I'm expected to approximate it. I'm guessing some sort of leading order term, but I don't know how to pick it out from the above expression.

Thanks!
Johnathan
 
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  • #2
Naturally, the leading terms for each of the potentials cancel. However, in order to get a non-zero contribution, you need to look to the next order (linear) in the quantity d/R rather than simply letting d->0.
 
  • #3
Ok I think I understand, but I'm not sure how to handle the 3rd power in the norm of the vector.
If it were a 2nd power, then defining x=d/r I would get something like (simplifying a little for readability):
1/|r[rhat]+dz[zhat]|^2=1/[r^2(1+2xz/r+x^2)]
which could then be approximated as
(1/r^2)*(1-2xz/r-x^2)

But in the original case, I'm not sure how to treat the vectors. Is something like the following makes sense?
|r-d/2z-hat|^3=|r^3[rhat]-3/2dzr^2[zhat]-3/2d^2z^2r[rhat]+d^3z^3/8[zhat]|

Or could it just be
|r-d/2z-hat|^3=(r^3-3/2dzr^2-3/2d^2z^2r+d^3z^3/8)
 
  • #4
Are you familiar with series expansions?
 
  • #5
We just started the course and have yet to cover series expansions...
 
  • #6
Just to be clear, I mean series expansions such as Taylor series expansions, not multipole expansions, which I suspect is something you would cover later.
 
  • #7
Not very thoroughly, but I'm familiar with common expansions such as sin/cos/e^x/ln(x), and with the idea of differentiating and dividing by the nth power etc.
I think I'm having more trouble with dealing with vector side of the problem, or maybe I'm just missing something...
 
  • #8
Ok I got it... It was just a matter of expressing r in cartesian coordinates, and transforming the absolute value in terms of root of square...

Thank you!
 

1. What is the formula for calculating the electric field due to 2 charged spheres?

The formula for calculating the electric field due to 2 charged spheres is given by:
E = k(q1/r1^2 + q2/r2^2)

2. How do the distances of the charged spheres affect the electric field?

The electric field is inversely proportional to the square of the distance between the two charged spheres. This means that as the distance increases, the electric field decreases.

3. Can the electric field due to 2 charged spheres be negative?

Yes, the electric field can be negative. This indicates the direction of the field is towards the negative charge, while a positive electric field indicates the direction is towards the positive charge.

4. What is the unit of measurement for electric field?

The unit of measurement for electric field is Newtons per Coulomb (N/C).

5. How do the magnitudes of the charges affect the electric field?

The electric field is directly proportional to the magnitude of the charges. This means that as the charges increase, the electric field also increases.

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