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E field at a distance

  • #1
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Homework Statement


consider 2 uniform static electric charge densities +/-[tex]\rho[/tex] located in the volumes defined by the 2 intercepting spheres of radius a. Charges are located in free space and [tex]\rho[/tex][tex]\delta[/tex] is a constant c.
The center of the spheres are separated by distance [tex]\delta[/tex]
The upper sphere has charge density +[tex]\rho[/tex] and lower has -[tex]\rho[/tex]

Show that E(P) in the region belonging to both spheres is constant and given by -[tex]\hat{z}[/tex]c/3[tex]\epsilon[/tex]0


Homework Equations





The Attempt at a Solution




Charge Q= [tex]\rho[/tex](4/3pi a3)

I tried for +ve charged sphere E(P) = (Q/(4pi[tex]\epsilon[/tex])R3)[ 3 R.d/R2 R-d]

I was able to get
E(P) = [tex]\rho[/tex]a3/3epsilonR3 [ 3 R [tex]\delta[/tex]/R2 R-[tex]\delta[/tex]]

I tried to use superposition and calculate E field due to each sphere and add up. But didn't get anywhere.
 

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Answers and Replies

  • #2
gabbagabbahey
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Charge Q= [tex]\rho[/tex](4/3pi a3)

I tried for +ve charged sphere E(P) = (Q/(4pi[tex]\epsilon[/tex])R3)[ 3 R.d/R2 R-d]
And how did you get this? If the positively charged sphere was at the origin instead, what would its electric field be inside and out (don't forget to include its direction)? What do you get when you shift your origin down [itex]\delta[/itex] units along the z-axis (i.e. [itex]\textbf{r}\to\textbf{r}-\delta\hat{\mathbf{z}}[/itex])?
 
  • #3
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I used the formula
E(P) = Kq/r2 [tex]\hat{r}[/tex]

[tex]\hat{r}[/tex] = r/r

r= (R-R')
here R'=[tex]\delta[/tex]/2 [tex]\hat{z}[/tex]
For a sphere at origin, R'=0.

After simplification and from Cheng's Field and EM waves eqn 3.26
E(P) = kq/R3 ( 3 R.d/R2 R-d]

(k=4piepsilon)
 
  • #4
gabbagabbahey
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I used the formula
E(P) = Kq/r2 [tex]\hat{r}[/tex]

[tex]\hat{r}[/tex] = r/r

r= (R-R')
here R'=[tex]\delta[/tex]/2 [tex]\hat{z}[/tex]

But that's the expression for the field due to a point charge. Use Gauss' Law instead (and start with the simple case of a sphere at the origin); you will find a different expression inside the sphere.

For a sphere at origin, R'=0.

After simplification and from Cheng's Field and EM waves eqn 3.26
E(P) = kq/R3 ( 3 R.d/R2 R-d]

(k=4piepsilon)
I don't have a copy of that text, but this equation doesn't look right at all.
 
  • #5
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Okay. I am trying to use just gauss law. But not sure about the R when the sphere is d/2 above the origin.
The first sphere is centered at R1, which is +d/2 above the origin and the second at R2, which is -d/2 from the origin.

E(P) for the first sphere is [tex]\rho[/tex]a3/3[tex]\epsilon[/tex]R12

for the second sphere E(P) is
- [tex]\rho[/tex]a3/3[tex]\epsilon[/tex]R22
I tried to express R1 and R2 in terms of R.
R1+d/2 =R
R2-d/2 =R

Adding the E fields I got
[tex]\rho[/tex]a3/3[tex]\epsilon[/tex] (2Rd/(R+d/2)2(R-d/2)2)
This is no where near what I want.
 
  • #6
gabbagabbahey
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E(P) for the first sphere is [tex]\rho[/tex]a3/3[tex]\epsilon[/tex]R12
No, it isn't. Show your work and explain each step in your application of Gauss' Law.
 
  • #7
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π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

Ok.
charge Q = ρ*volume
Q= ρ 4pi a3/3

E field at P is

[tex]\oint[/tex]E.ds = Q/ε

E.Area = Q/ε

E = Q/A.ε

Area of sphere with radius R1 is 4piR12
R1 is the distance from the center of sphere 1 to point P.

E= ρ 4pia3/3ε*4pi*a2


E= ρ a3/3εR12
 
  • #8
gabbagabbahey
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Gauss' Law involves the charge enclosed by your Gaussian surface. For R1<a, that will not be the same as the total charge of the sphere.
 
  • #9
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R1 is not < a. It's >a. It's the sphere with radius comparable to R. R1 is the distance from the center of the 1st sphere to the point P.
 
  • #10
gabbagabbahey
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You are asked to find the electric field at a point "belonging to both spheres". That means the point will be inside each sphere and so R1<a
 
  • #11
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Well, the problem asks to find E(P) in the region belonging to both spheres. Since it's E(P), I thought it was E field at point P.
If I take the region belonging to both spheres,
Q will be ρ4pi (a-δ/2)^3/3ε
 
  • #12
gabbagabbahey
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Well, the problem asks to find E(P) in the region belonging to both spheres. Since it's E(P), I thought it was E field at point P.
If I take the region belonging to both spheres,
Q will be ρ4pi (a-δ/2)^3/3ε
That looks better! What does that give you for the field due to the positive sphere? Apply the same method to the negative sphere and then usde the superposition principle.
 
  • #13
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π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

Got close, but not the answer.

For the positive sphere

E1 will be

E1 (4pia2) = ρ4pi (a-δ/2)^3/3ε

E1 = ρ(a-δ/2)^3/3εa2

Similarly, for the negatively charged sphere

E2 = -ρ(a+δ/2)^3/3εa2

Combining,

E = (ρ/3εa) 2δ

But the required answer is E= -[tex]\hat{z}[/tex] (ρδ)/3ε

I can't figure out where I am going wrong.
 
  • #14
gabbagabbahey
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Why are you dividing by [itex]4\pi a^2[/itex]? That's not really the surface area of your Gaussian surface is it?
 
  • #15
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Why are you dividing by [itex]4\pi a^2[/itex]? That's not really the surface area of your Gaussian surface is it?
Isn't it? When calculating E only for the +ve sphere, the area is 4pia2.
 
  • #16
gabbagabbahey
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Isn't it? When calculating E only for the +ve sphere, the area is 4pia2.
The radius of your Gaussian sphere should be the distance from the center to the point you are looking at. If you are looking at a point inside the sphere, that radius will be lees than [itex]a[/itex].
 
  • #17
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I agree that radius will be less than a. But I am trying to use superposition theorem. I should be able to get the answer by calculating E for reach sphere and then adding, correct?
Why should I consider the area of overlap?
 
  • #18
gabbagabbahey
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You seem to be confusing yourself here. Forget about the actual problem at hand for a moment and calculate the field due to a single uniformly charged sphere, centered at the origin. What is the field inside the sphere? What is the field outside? Show your calculations.
 
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